Proof that $lim_{n to infty} n[log (n+1)-log(n)]=1$












0












$begingroup$


Can someone explain me why $lim_{n to infty} n[log (n+1)-log(n)]=1$? Isn't that an indeterminate form? I mean, since $lim_{n to infty} n = infty$ and $lim_{n to infty} [log (n+1)-log(n)] = 0$ then $lim_{n to infty} a_n b_n=0 cdotinfty$?



I'm just starting with sequences, and I have no idea what to do.






calculus sequences-and-series limits







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$endgroup$








  • 1




    $begingroup$
    You can only say that $lim a_n b_n = lim a_n cdot lim b_n$, if both limits exist. If one is $infty$, as in your example, the rule doesn't apply.
    $endgroup$
    – Viktor Glombik
    Dec 23 '18 at 22:12










  • $begingroup$
    You may think of $log n$ as $int_{1}^{infty}frac1x dx$.
    $endgroup$
    – Leila
    Dec 24 '18 at 15:09
















0












$begingroup$


Can someone explain me why $lim_{n to infty} n[log (n+1)-log(n)]=1$? Isn't that an indeterminate form? I mean, since $lim_{n to infty} n = infty$ and $lim_{n to infty} [log (n+1)-log(n)] = 0$ then $lim_{n to infty} a_n b_n=0 cdotinfty$?



I'm just starting with sequences, and I have no idea what to do.






calculus sequences-and-series limits







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    You can only say that $lim a_n b_n = lim a_n cdot lim b_n$, if both limits exist. If one is $infty$, as in your example, the rule doesn't apply.
    $endgroup$
    – Viktor Glombik
    Dec 23 '18 at 22:12










  • $begingroup$
    You may think of $log n$ as $int_{1}^{infty}frac1x dx$.
    $endgroup$
    – Leila
    Dec 24 '18 at 15:09














0












0








0





$begingroup$


Can someone explain me why $lim_{n to infty} n[log (n+1)-log(n)]=1$? Isn't that an indeterminate form? I mean, since $lim_{n to infty} n = infty$ and $lim_{n to infty} [log (n+1)-log(n)] = 0$ then $lim_{n to infty} a_n b_n=0 cdotinfty$?



I'm just starting with sequences, and I have no idea what to do.






calculus sequences-and-series limits







share|cite|improve this question









$endgroup$




Can someone explain me why $lim_{n to infty} n[log (n+1)-log(n)]=1$? Isn't that an indeterminate form? I mean, since $lim_{n to infty} n = infty$ and $lim_{n to infty} [log (n+1)-log(n)] = 0$ then $lim_{n to infty} a_n b_n=0 cdotinfty$?



I'm just starting with sequences, and I have no idea what to do.






calculus sequences-and-series limits




calculus sequences-and-series limits






share|cite|improve this question













share|cite|improve this question











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asked Dec 23 '18 at 22:09









parishiltonparishilton

19310




19310








  • 1




    $begingroup$
    You can only say that $lim a_n b_n = lim a_n cdot lim b_n$, if both limits exist. If one is $infty$, as in your example, the rule doesn't apply.
    $endgroup$
    – Viktor Glombik
    Dec 23 '18 at 22:12










  • $begingroup$
    You may think of $log n$ as $int_{1}^{infty}frac1x dx$.
    $endgroup$
    – Leila
    Dec 24 '18 at 15:09














  • 1




    $begingroup$
    You can only say that $lim a_n b_n = lim a_n cdot lim b_n$, if both limits exist. If one is $infty$, as in your example, the rule doesn't apply.
    $endgroup$
    – Viktor Glombik
    Dec 23 '18 at 22:12










  • $begingroup$
    You may think of $log n$ as $int_{1}^{infty}frac1x dx$.
    $endgroup$
    – Leila
    Dec 24 '18 at 15:09








1




1




$begingroup$
You can only say that $lim a_n b_n = lim a_n cdot lim b_n$, if both limits exist. If one is $infty$, as in your example, the rule doesn't apply.
$endgroup$
– Viktor Glombik
Dec 23 '18 at 22:12




$begingroup$
You can only say that $lim a_n b_n = lim a_n cdot lim b_n$, if both limits exist. If one is $infty$, as in your example, the rule doesn't apply.
$endgroup$
– Viktor Glombik
Dec 23 '18 at 22:12












$begingroup$
You may think of $log n$ as $int_{1}^{infty}frac1x dx$.
$endgroup$
– Leila
Dec 24 '18 at 15:09




$begingroup$
You may think of $log n$ as $int_{1}^{infty}frac1x dx$.
$endgroup$
– Leila
Dec 24 '18 at 15:09










4 Answers
4






active

oldest

votes


















9












$begingroup$

As an alternative without L'Hopital, but rather using the very well known limit:
$$
begin{align}
lim_{ntoinfty}left(ncdotleft(ln(n+1) - ln nright)right) &= lim_{ntoinfty}left(ncdotlnleft(frac{n+1}{n}right)right) \
&= lim_{ntoinfty} lnleft(frac{n+1}{n}right)^n\
&= ln lim_{ntoinfty}left(1+frac{1}{n}right)^n\
&= ln e \
&= 1
end{align}
$$



I'm assuming you used $log x$ for $log_ex$ or simply $ln x$.






share|cite|improve this answer











$endgroup$





















    4












    $begingroup$

    If you already know the rule of L'HOSPITAL, the calculation could go as follows
    begin{align*}
    L
    := lim_{n to infty} n cdot log left( frac{n + 1}{n} right)
    = lim_{n to infty} frac{log left( frac{n + 1}{n} right)}{frac{1}{n}}
    end{align*}
    (here I am only using logarithm and fraction properties)



    Now this in an indeterminate form ''$frac{0}{0}$'', so we can apply the above mentioned rule by differentiating the numerator and the denominator:
    begin{align*}
    L
    = lim_{n to infty} frac{frac{n}{n + 1} left( frac{1}{n} -frac{n + 1}{n^2} right)}{- frac{1}{n^2}}
    = lim_{n to infty} frac{n}{n + 1}
    = 1.
    end{align*}
    This follows because
    begin{equation*}
    frac{d}{dx} frac{1}{x}
    = frac{d}{dx} x^{-1}
    = - x^{-2}
    end{equation*}
    and
    begin{equation*}
    frac{d}{dx} logleft(frac{x + 1}{x} right)
    = frac{d}{dx} log(x + 1) - frac{d}{dx} log(x)
    = frac{1}{x + 1} - frac{1}{x}
    = frac{x}{x + 1} left( frac{1}{x} -frac{x + 1}{x^2} right)
    end{equation*}






    share|cite|improve this answer









    $endgroup$





















      4












      $begingroup$

      It's because it can rewritten as
      $$nlogfrac{n+1}n=nlogBigl(1+frac1nBigr)=frac{logBigl(1+dfrac1nBigr)}{dfrac1n}$$
      Set $u=dfrac1n$. This expression becomes $dfrac{log(1+u)}u$, and it is standard from high school that the limit of this quotient as $uto 0$ is$;bigl(log(1+u)bigr)'_{u=0},:$ i.e. $:1$.






      share|cite|improve this answer









      $endgroup$





















        2












        $begingroup$

        You could even go beyond the limit. Write
        $$a_n= n[log (n+1)-log(n)]=n logleft(frac{n+1}n right)=n logleft(1+frac{1}n right)$$
        Now, remembering that for small $x$
        $$log(1+x)=x-frac{x^2}{2}+frac{x^3}{3}+Oleft(x^4right)$$ make $x=frac 1 n$ to get
        $$a_n=nleft(frac{1}{n}-frac{1}{2 n^2}+frac{1}{3
        n^3}+Oleft(frac{1}{n^4}right)right)=1-frac{1}{2 n}+frac{1}{3
        n^2}+Oleft(frac{1}{n^3}right)$$



        Just try it for $n=10$. You will get
        $$a_{10}=10 ,log left(frac{11}{10}right)approx 0.953102$$ while the above expansion would give
        $$a_{10}simeq frac{143}{150}approx 0.953333$$






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          This time faster than @gimusi :)
          $endgroup$
          – roman
          Dec 24 '18 at 11:38








        • 1




          $begingroup$
          @roman. For once !! By the way, Merry Xmas
          $endgroup$
          – Claude Leibovici
          Dec 24 '18 at 11:40











        Your Answer





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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        9












        $begingroup$

        As an alternative without L'Hopital, but rather using the very well known limit:
        $$
        begin{align}
        lim_{ntoinfty}left(ncdotleft(ln(n+1) - ln nright)right) &= lim_{ntoinfty}left(ncdotlnleft(frac{n+1}{n}right)right) \
        &= lim_{ntoinfty} lnleft(frac{n+1}{n}right)^n\
        &= ln lim_{ntoinfty}left(1+frac{1}{n}right)^n\
        &= ln e \
        &= 1
        end{align}
        $$



        I'm assuming you used $log x$ for $log_ex$ or simply $ln x$.






        share|cite|improve this answer











        $endgroup$


















          9












          $begingroup$

          As an alternative without L'Hopital, but rather using the very well known limit:
          $$
          begin{align}
          lim_{ntoinfty}left(ncdotleft(ln(n+1) - ln nright)right) &= lim_{ntoinfty}left(ncdotlnleft(frac{n+1}{n}right)right) \
          &= lim_{ntoinfty} lnleft(frac{n+1}{n}right)^n\
          &= ln lim_{ntoinfty}left(1+frac{1}{n}right)^n\
          &= ln e \
          &= 1
          end{align}
          $$



          I'm assuming you used $log x$ for $log_ex$ or simply $ln x$.






          share|cite|improve this answer











          $endgroup$
















            9












            9








            9





            $begingroup$

            As an alternative without L'Hopital, but rather using the very well known limit:
            $$
            begin{align}
            lim_{ntoinfty}left(ncdotleft(ln(n+1) - ln nright)right) &= lim_{ntoinfty}left(ncdotlnleft(frac{n+1}{n}right)right) \
            &= lim_{ntoinfty} lnleft(frac{n+1}{n}right)^n\
            &= ln lim_{ntoinfty}left(1+frac{1}{n}right)^n\
            &= ln e \
            &= 1
            end{align}
            $$



            I'm assuming you used $log x$ for $log_ex$ or simply $ln x$.






            share|cite|improve this answer











            $endgroup$



            As an alternative without L'Hopital, but rather using the very well known limit:
            $$
            begin{align}
            lim_{ntoinfty}left(ncdotleft(ln(n+1) - ln nright)right) &= lim_{ntoinfty}left(ncdotlnleft(frac{n+1}{n}right)right) \
            &= lim_{ntoinfty} lnleft(frac{n+1}{n}right)^n\
            &= ln lim_{ntoinfty}left(1+frac{1}{n}right)^n\
            &= ln e \
            &= 1
            end{align}
            $$



            I'm assuming you used $log x$ for $log_ex$ or simply $ln x$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 23 '18 at 22:29









            Viktor Glombik

            9381527




            9381527










            answered Dec 23 '18 at 22:19









            romanroman

            2,28421224




            2,28421224























                4












                $begingroup$

                If you already know the rule of L'HOSPITAL, the calculation could go as follows
                begin{align*}
                L
                := lim_{n to infty} n cdot log left( frac{n + 1}{n} right)
                = lim_{n to infty} frac{log left( frac{n + 1}{n} right)}{frac{1}{n}}
                end{align*}
                (here I am only using logarithm and fraction properties)



                Now this in an indeterminate form ''$frac{0}{0}$'', so we can apply the above mentioned rule by differentiating the numerator and the denominator:
                begin{align*}
                L
                = lim_{n to infty} frac{frac{n}{n + 1} left( frac{1}{n} -frac{n + 1}{n^2} right)}{- frac{1}{n^2}}
                = lim_{n to infty} frac{n}{n + 1}
                = 1.
                end{align*}
                This follows because
                begin{equation*}
                frac{d}{dx} frac{1}{x}
                = frac{d}{dx} x^{-1}
                = - x^{-2}
                end{equation*}
                and
                begin{equation*}
                frac{d}{dx} logleft(frac{x + 1}{x} right)
                = frac{d}{dx} log(x + 1) - frac{d}{dx} log(x)
                = frac{1}{x + 1} - frac{1}{x}
                = frac{x}{x + 1} left( frac{1}{x} -frac{x + 1}{x^2} right)
                end{equation*}






                share|cite|improve this answer









                $endgroup$


















                  4












                  $begingroup$

                  If you already know the rule of L'HOSPITAL, the calculation could go as follows
                  begin{align*}
                  L
                  := lim_{n to infty} n cdot log left( frac{n + 1}{n} right)
                  = lim_{n to infty} frac{log left( frac{n + 1}{n} right)}{frac{1}{n}}
                  end{align*}
                  (here I am only using logarithm and fraction properties)



                  Now this in an indeterminate form ''$frac{0}{0}$'', so we can apply the above mentioned rule by differentiating the numerator and the denominator:
                  begin{align*}
                  L
                  = lim_{n to infty} frac{frac{n}{n + 1} left( frac{1}{n} -frac{n + 1}{n^2} right)}{- frac{1}{n^2}}
                  = lim_{n to infty} frac{n}{n + 1}
                  = 1.
                  end{align*}
                  This follows because
                  begin{equation*}
                  frac{d}{dx} frac{1}{x}
                  = frac{d}{dx} x^{-1}
                  = - x^{-2}
                  end{equation*}
                  and
                  begin{equation*}
                  frac{d}{dx} logleft(frac{x + 1}{x} right)
                  = frac{d}{dx} log(x + 1) - frac{d}{dx} log(x)
                  = frac{1}{x + 1} - frac{1}{x}
                  = frac{x}{x + 1} left( frac{1}{x} -frac{x + 1}{x^2} right)
                  end{equation*}






                  share|cite|improve this answer









                  $endgroup$
















                    4












                    4








                    4





                    $begingroup$

                    If you already know the rule of L'HOSPITAL, the calculation could go as follows
                    begin{align*}
                    L
                    := lim_{n to infty} n cdot log left( frac{n + 1}{n} right)
                    = lim_{n to infty} frac{log left( frac{n + 1}{n} right)}{frac{1}{n}}
                    end{align*}
                    (here I am only using logarithm and fraction properties)



                    Now this in an indeterminate form ''$frac{0}{0}$'', so we can apply the above mentioned rule by differentiating the numerator and the denominator:
                    begin{align*}
                    L
                    = lim_{n to infty} frac{frac{n}{n + 1} left( frac{1}{n} -frac{n + 1}{n^2} right)}{- frac{1}{n^2}}
                    = lim_{n to infty} frac{n}{n + 1}
                    = 1.
                    end{align*}
                    This follows because
                    begin{equation*}
                    frac{d}{dx} frac{1}{x}
                    = frac{d}{dx} x^{-1}
                    = - x^{-2}
                    end{equation*}
                    and
                    begin{equation*}
                    frac{d}{dx} logleft(frac{x + 1}{x} right)
                    = frac{d}{dx} log(x + 1) - frac{d}{dx} log(x)
                    = frac{1}{x + 1} - frac{1}{x}
                    = frac{x}{x + 1} left( frac{1}{x} -frac{x + 1}{x^2} right)
                    end{equation*}






                    share|cite|improve this answer









                    $endgroup$



                    If you already know the rule of L'HOSPITAL, the calculation could go as follows
                    begin{align*}
                    L
                    := lim_{n to infty} n cdot log left( frac{n + 1}{n} right)
                    = lim_{n to infty} frac{log left( frac{n + 1}{n} right)}{frac{1}{n}}
                    end{align*}
                    (here I am only using logarithm and fraction properties)



                    Now this in an indeterminate form ''$frac{0}{0}$'', so we can apply the above mentioned rule by differentiating the numerator and the denominator:
                    begin{align*}
                    L
                    = lim_{n to infty} frac{frac{n}{n + 1} left( frac{1}{n} -frac{n + 1}{n^2} right)}{- frac{1}{n^2}}
                    = lim_{n to infty} frac{n}{n + 1}
                    = 1.
                    end{align*}
                    This follows because
                    begin{equation*}
                    frac{d}{dx} frac{1}{x}
                    = frac{d}{dx} x^{-1}
                    = - x^{-2}
                    end{equation*}
                    and
                    begin{equation*}
                    frac{d}{dx} logleft(frac{x + 1}{x} right)
                    = frac{d}{dx} log(x + 1) - frac{d}{dx} log(x)
                    = frac{1}{x + 1} - frac{1}{x}
                    = frac{x}{x + 1} left( frac{1}{x} -frac{x + 1}{x^2} right)
                    end{equation*}







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 23 '18 at 22:17









                    Viktor GlombikViktor Glombik

                    9381527




                    9381527























                        4












                        $begingroup$

                        It's because it can rewritten as
                        $$nlogfrac{n+1}n=nlogBigl(1+frac1nBigr)=frac{logBigl(1+dfrac1nBigr)}{dfrac1n}$$
                        Set $u=dfrac1n$. This expression becomes $dfrac{log(1+u)}u$, and it is standard from high school that the limit of this quotient as $uto 0$ is$;bigl(log(1+u)bigr)'_{u=0},:$ i.e. $:1$.






                        share|cite|improve this answer









                        $endgroup$


















                          4












                          $begingroup$

                          It's because it can rewritten as
                          $$nlogfrac{n+1}n=nlogBigl(1+frac1nBigr)=frac{logBigl(1+dfrac1nBigr)}{dfrac1n}$$
                          Set $u=dfrac1n$. This expression becomes $dfrac{log(1+u)}u$, and it is standard from high school that the limit of this quotient as $uto 0$ is$;bigl(log(1+u)bigr)'_{u=0},:$ i.e. $:1$.






                          share|cite|improve this answer









                          $endgroup$
















                            4












                            4








                            4





                            $begingroup$

                            It's because it can rewritten as
                            $$nlogfrac{n+1}n=nlogBigl(1+frac1nBigr)=frac{logBigl(1+dfrac1nBigr)}{dfrac1n}$$
                            Set $u=dfrac1n$. This expression becomes $dfrac{log(1+u)}u$, and it is standard from high school that the limit of this quotient as $uto 0$ is$;bigl(log(1+u)bigr)'_{u=0},:$ i.e. $:1$.






                            share|cite|improve this answer









                            $endgroup$



                            It's because it can rewritten as
                            $$nlogfrac{n+1}n=nlogBigl(1+frac1nBigr)=frac{logBigl(1+dfrac1nBigr)}{dfrac1n}$$
                            Set $u=dfrac1n$. This expression becomes $dfrac{log(1+u)}u$, and it is standard from high school that the limit of this quotient as $uto 0$ is$;bigl(log(1+u)bigr)'_{u=0},:$ i.e. $:1$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 23 '18 at 22:18









                            BernardBernard

                            121k740116




                            121k740116























                                2












                                $begingroup$

                                You could even go beyond the limit. Write
                                $$a_n= n[log (n+1)-log(n)]=n logleft(frac{n+1}n right)=n logleft(1+frac{1}n right)$$
                                Now, remembering that for small $x$
                                $$log(1+x)=x-frac{x^2}{2}+frac{x^3}{3}+Oleft(x^4right)$$ make $x=frac 1 n$ to get
                                $$a_n=nleft(frac{1}{n}-frac{1}{2 n^2}+frac{1}{3
                                n^3}+Oleft(frac{1}{n^4}right)right)=1-frac{1}{2 n}+frac{1}{3
                                n^2}+Oleft(frac{1}{n^3}right)$$



                                Just try it for $n=10$. You will get
                                $$a_{10}=10 ,log left(frac{11}{10}right)approx 0.953102$$ while the above expansion would give
                                $$a_{10}simeq frac{143}{150}approx 0.953333$$






                                share|cite|improve this answer









                                $endgroup$













                                • $begingroup$
                                  This time faster than @gimusi :)
                                  $endgroup$
                                  – roman
                                  Dec 24 '18 at 11:38








                                • 1




                                  $begingroup$
                                  @roman. For once !! By the way, Merry Xmas
                                  $endgroup$
                                  – Claude Leibovici
                                  Dec 24 '18 at 11:40
















                                2












                                $begingroup$

                                You could even go beyond the limit. Write
                                $$a_n= n[log (n+1)-log(n)]=n logleft(frac{n+1}n right)=n logleft(1+frac{1}n right)$$
                                Now, remembering that for small $x$
                                $$log(1+x)=x-frac{x^2}{2}+frac{x^3}{3}+Oleft(x^4right)$$ make $x=frac 1 n$ to get
                                $$a_n=nleft(frac{1}{n}-frac{1}{2 n^2}+frac{1}{3
                                n^3}+Oleft(frac{1}{n^4}right)right)=1-frac{1}{2 n}+frac{1}{3
                                n^2}+Oleft(frac{1}{n^3}right)$$



                                Just try it for $n=10$. You will get
                                $$a_{10}=10 ,log left(frac{11}{10}right)approx 0.953102$$ while the above expansion would give
                                $$a_{10}simeq frac{143}{150}approx 0.953333$$






                                share|cite|improve this answer









                                $endgroup$













                                • $begingroup$
                                  This time faster than @gimusi :)
                                  $endgroup$
                                  – roman
                                  Dec 24 '18 at 11:38








                                • 1




                                  $begingroup$
                                  @roman. For once !! By the way, Merry Xmas
                                  $endgroup$
                                  – Claude Leibovici
                                  Dec 24 '18 at 11:40














                                2












                                2








                                2





                                $begingroup$

                                You could even go beyond the limit. Write
                                $$a_n= n[log (n+1)-log(n)]=n logleft(frac{n+1}n right)=n logleft(1+frac{1}n right)$$
                                Now, remembering that for small $x$
                                $$log(1+x)=x-frac{x^2}{2}+frac{x^3}{3}+Oleft(x^4right)$$ make $x=frac 1 n$ to get
                                $$a_n=nleft(frac{1}{n}-frac{1}{2 n^2}+frac{1}{3
                                n^3}+Oleft(frac{1}{n^4}right)right)=1-frac{1}{2 n}+frac{1}{3
                                n^2}+Oleft(frac{1}{n^3}right)$$



                                Just try it for $n=10$. You will get
                                $$a_{10}=10 ,log left(frac{11}{10}right)approx 0.953102$$ while the above expansion would give
                                $$a_{10}simeq frac{143}{150}approx 0.953333$$






                                share|cite|improve this answer









                                $endgroup$



                                You could even go beyond the limit. Write
                                $$a_n= n[log (n+1)-log(n)]=n logleft(frac{n+1}n right)=n logleft(1+frac{1}n right)$$
                                Now, remembering that for small $x$
                                $$log(1+x)=x-frac{x^2}{2}+frac{x^3}{3}+Oleft(x^4right)$$ make $x=frac 1 n$ to get
                                $$a_n=nleft(frac{1}{n}-frac{1}{2 n^2}+frac{1}{3
                                n^3}+Oleft(frac{1}{n^4}right)right)=1-frac{1}{2 n}+frac{1}{3
                                n^2}+Oleft(frac{1}{n^3}right)$$



                                Just try it for $n=10$. You will get
                                $$a_{10}=10 ,log left(frac{11}{10}right)approx 0.953102$$ while the above expansion would give
                                $$a_{10}simeq frac{143}{150}approx 0.953333$$







                                share|cite|improve this answer












                                share|cite|improve this answer



                                share|cite|improve this answer










                                answered Dec 24 '18 at 11:37









                                Claude LeiboviciClaude Leibovici

                                122k1157134




                                122k1157134












                                • $begingroup$
                                  This time faster than @gimusi :)
                                  $endgroup$
                                  – roman
                                  Dec 24 '18 at 11:38








                                • 1




                                  $begingroup$
                                  @roman. For once !! By the way, Merry Xmas
                                  $endgroup$
                                  – Claude Leibovici
                                  Dec 24 '18 at 11:40


















                                • $begingroup$
                                  This time faster than @gimusi :)
                                  $endgroup$
                                  – roman
                                  Dec 24 '18 at 11:38








                                • 1




                                  $begingroup$
                                  @roman. For once !! By the way, Merry Xmas
                                  $endgroup$
                                  – Claude Leibovici
                                  Dec 24 '18 at 11:40
















                                $begingroup$
                                This time faster than @gimusi :)
                                $endgroup$
                                – roman
                                Dec 24 '18 at 11:38






                                $begingroup$
                                This time faster than @gimusi :)
                                $endgroup$
                                – roman
                                Dec 24 '18 at 11:38






                                1




                                1




                                $begingroup$
                                @roman. For once !! By the way, Merry Xmas
                                $endgroup$
                                – Claude Leibovici
                                Dec 24 '18 at 11:40




                                $begingroup$
                                @roman. For once !! By the way, Merry Xmas
                                $endgroup$
                                – Claude Leibovici
                                Dec 24 '18 at 11:40


















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