Proving that the sum of the difference of square roots of partial sums diverges












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Let $(a_n)_{n∈N}$ be a sequence of positive numbers which tends to zero but such that $sum_{n=1}^infty a_n$ diverges. Let $(A_n)_{n∈N}$ be the sequence of partial sums $A_n=sum_{k=1}^n a_k$,and let $b_{n+1}=sqrt{A_{n+1}}−sqrt{A_{n}}$. Show that $lim_{n→∞}frac{b_n}{a_n}= 0$ but that $sum b_n$ is still divergent.




Question 1:



I was able to answer the first part, since



$$frac{b_{n+1}}{a_{n+1}} = frac{sqrt{A_{n+1}}−sqrt{A_{n}}}{a_{n+1}} = frac{A_{n+1}-A_n}{a_{n+1}(sqrt{A_{n+1}}+sqrt{A_{n}})} = frac{1}{sqrt{A_{n+1}}+sqrt{A_{n}}}$$ which converges to $0$ since $sqrt{A_n}$ tends to infinity.



The second part, I have no idea about. Apparently $$sum b_k = sqrt{A_n}$$ Why is this true?



Question 2: It is also said that




In this sense there is no ‘smallest’ divergent series, and one can similarly show that there is no ‘largest’ convergent one.




I don't see why this follows. What about our findings conveys these ideas?






real-analysis sequences-and-series convergence







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    For question 1) just write down $b_1+b_2+cdots +b_N$ and observe that lots of terms cancel out.
    $endgroup$
    – Kavi Rama Murthy
    Oct 17 '18 at 23:21
















0












$begingroup$



Let $(a_n)_{n∈N}$ be a sequence of positive numbers which tends to zero but such that $sum_{n=1}^infty a_n$ diverges. Let $(A_n)_{n∈N}$ be the sequence of partial sums $A_n=sum_{k=1}^n a_k$,and let $b_{n+1}=sqrt{A_{n+1}}−sqrt{A_{n}}$. Show that $lim_{n→∞}frac{b_n}{a_n}= 0$ but that $sum b_n$ is still divergent.




Question 1:



I was able to answer the first part, since



$$frac{b_{n+1}}{a_{n+1}} = frac{sqrt{A_{n+1}}−sqrt{A_{n}}}{a_{n+1}} = frac{A_{n+1}-A_n}{a_{n+1}(sqrt{A_{n+1}}+sqrt{A_{n}})} = frac{1}{sqrt{A_{n+1}}+sqrt{A_{n}}}$$ which converges to $0$ since $sqrt{A_n}$ tends to infinity.



The second part, I have no idea about. Apparently $$sum b_k = sqrt{A_n}$$ Why is this true?



Question 2: It is also said that




In this sense there is no ‘smallest’ divergent series, and one can similarly show that there is no ‘largest’ convergent one.




I don't see why this follows. What about our findings conveys these ideas?






real-analysis sequences-and-series convergence







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    For question 1) just write down $b_1+b_2+cdots +b_N$ and observe that lots of terms cancel out.
    $endgroup$
    – Kavi Rama Murthy
    Oct 17 '18 at 23:21














0












0








0





$begingroup$



Let $(a_n)_{n∈N}$ be a sequence of positive numbers which tends to zero but such that $sum_{n=1}^infty a_n$ diverges. Let $(A_n)_{n∈N}$ be the sequence of partial sums $A_n=sum_{k=1}^n a_k$,and let $b_{n+1}=sqrt{A_{n+1}}−sqrt{A_{n}}$. Show that $lim_{n→∞}frac{b_n}{a_n}= 0$ but that $sum b_n$ is still divergent.




Question 1:



I was able to answer the first part, since



$$frac{b_{n+1}}{a_{n+1}} = frac{sqrt{A_{n+1}}−sqrt{A_{n}}}{a_{n+1}} = frac{A_{n+1}-A_n}{a_{n+1}(sqrt{A_{n+1}}+sqrt{A_{n}})} = frac{1}{sqrt{A_{n+1}}+sqrt{A_{n}}}$$ which converges to $0$ since $sqrt{A_n}$ tends to infinity.



The second part, I have no idea about. Apparently $$sum b_k = sqrt{A_n}$$ Why is this true?



Question 2: It is also said that




In this sense there is no ‘smallest’ divergent series, and one can similarly show that there is no ‘largest’ convergent one.




I don't see why this follows. What about our findings conveys these ideas?






real-analysis sequences-and-series convergence







share|cite|improve this question









$endgroup$





Let $(a_n)_{n∈N}$ be a sequence of positive numbers which tends to zero but such that $sum_{n=1}^infty a_n$ diverges. Let $(A_n)_{n∈N}$ be the sequence of partial sums $A_n=sum_{k=1}^n a_k$,and let $b_{n+1}=sqrt{A_{n+1}}−sqrt{A_{n}}$. Show that $lim_{n→∞}frac{b_n}{a_n}= 0$ but that $sum b_n$ is still divergent.




Question 1:



I was able to answer the first part, since



$$frac{b_{n+1}}{a_{n+1}} = frac{sqrt{A_{n+1}}−sqrt{A_{n}}}{a_{n+1}} = frac{A_{n+1}-A_n}{a_{n+1}(sqrt{A_{n+1}}+sqrt{A_{n}})} = frac{1}{sqrt{A_{n+1}}+sqrt{A_{n}}}$$ which converges to $0$ since $sqrt{A_n}$ tends to infinity.



The second part, I have no idea about. Apparently $$sum b_k = sqrt{A_n}$$ Why is this true?



Question 2: It is also said that




In this sense there is no ‘smallest’ divergent series, and one can similarly show that there is no ‘largest’ convergent one.




I don't see why this follows. What about our findings conveys these ideas?






real-analysis sequences-and-series convergence




real-analysis sequences-and-series convergence






share|cite|improve this question













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asked Oct 17 '18 at 23:09









Tiwa AinaTiwa Aina

2,686421




2,686421








  • 1




    $begingroup$
    For question 1) just write down $b_1+b_2+cdots +b_N$ and observe that lots of terms cancel out.
    $endgroup$
    – Kavi Rama Murthy
    Oct 17 '18 at 23:21














  • 1




    $begingroup$
    For question 1) just write down $b_1+b_2+cdots +b_N$ and observe that lots of terms cancel out.
    $endgroup$
    – Kavi Rama Murthy
    Oct 17 '18 at 23:21








1




1




$begingroup$
For question 1) just write down $b_1+b_2+cdots +b_N$ and observe that lots of terms cancel out.
$endgroup$
– Kavi Rama Murthy
Oct 17 '18 at 23:21




$begingroup$
For question 1) just write down $b_1+b_2+cdots +b_N$ and observe that lots of terms cancel out.
$endgroup$
– Kavi Rama Murthy
Oct 17 '18 at 23:21










1 Answer
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We have$$sum_{k=1}^{n}b_{k+1}{=sum_{k=1}^{n}sqrt{A_{k+1}}-sqrt{A_k}\=sqrt{A_{2}}-sqrt{A_1}\+sqrt{A_{3}}-sqrt{2}\+sqrt{A_{4}}-sqrt{A_3}\+cdots\+sqrt{A_{n+1}}-sqrt{A_n}}\=sqrt{A_{n+1}}-sqrt{A_1}$$according to telescopic rule (https://en.wikipedia.org/wiki/Telescoping_series) therefore $sum_{k=1}^{n}b_{k+1}$ diverges since $sqrt{A_{n+1}}-sqrt{A_1}$ diverges.






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    $begingroup$

    We have$$sum_{k=1}^{n}b_{k+1}{=sum_{k=1}^{n}sqrt{A_{k+1}}-sqrt{A_k}\=sqrt{A_{2}}-sqrt{A_1}\+sqrt{A_{3}}-sqrt{2}\+sqrt{A_{4}}-sqrt{A_3}\+cdots\+sqrt{A_{n+1}}-sqrt{A_n}}\=sqrt{A_{n+1}}-sqrt{A_1}$$according to telescopic rule (https://en.wikipedia.org/wiki/Telescoping_series) therefore $sum_{k=1}^{n}b_{k+1}$ diverges since $sqrt{A_{n+1}}-sqrt{A_1}$ diverges.






    share|cite|improve this answer









    $endgroup$


















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      $begingroup$

      We have$$sum_{k=1}^{n}b_{k+1}{=sum_{k=1}^{n}sqrt{A_{k+1}}-sqrt{A_k}\=sqrt{A_{2}}-sqrt{A_1}\+sqrt{A_{3}}-sqrt{2}\+sqrt{A_{4}}-sqrt{A_3}\+cdots\+sqrt{A_{n+1}}-sqrt{A_n}}\=sqrt{A_{n+1}}-sqrt{A_1}$$according to telescopic rule (https://en.wikipedia.org/wiki/Telescoping_series) therefore $sum_{k=1}^{n}b_{k+1}$ diverges since $sqrt{A_{n+1}}-sqrt{A_1}$ diverges.






      share|cite|improve this answer









      $endgroup$
















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        0








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        $begingroup$

        We have$$sum_{k=1}^{n}b_{k+1}{=sum_{k=1}^{n}sqrt{A_{k+1}}-sqrt{A_k}\=sqrt{A_{2}}-sqrt{A_1}\+sqrt{A_{3}}-sqrt{2}\+sqrt{A_{4}}-sqrt{A_3}\+cdots\+sqrt{A_{n+1}}-sqrt{A_n}}\=sqrt{A_{n+1}}-sqrt{A_1}$$according to telescopic rule (https://en.wikipedia.org/wiki/Telescoping_series) therefore $sum_{k=1}^{n}b_{k+1}$ diverges since $sqrt{A_{n+1}}-sqrt{A_1}$ diverges.






        share|cite|improve this answer









        $endgroup$



        We have$$sum_{k=1}^{n}b_{k+1}{=sum_{k=1}^{n}sqrt{A_{k+1}}-sqrt{A_k}\=sqrt{A_{2}}-sqrt{A_1}\+sqrt{A_{3}}-sqrt{2}\+sqrt{A_{4}}-sqrt{A_3}\+cdots\+sqrt{A_{n+1}}-sqrt{A_n}}\=sqrt{A_{n+1}}-sqrt{A_1}$$according to telescopic rule (https://en.wikipedia.org/wiki/Telescoping_series) therefore $sum_{k=1}^{n}b_{k+1}$ diverges since $sqrt{A_{n+1}}-sqrt{A_1}$ diverges.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 30 '18 at 19:26









        Mostafa AyazMostafa Ayaz

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        14.8k3938






























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