Summation equivalence












1












$begingroup$


I am having trouble seeing how this summation equivalence holds true:



$sum_{i=0}^infty x^i$ = $frac{1}{1-x}$ if |x| < 1



The only thing I can see where there would be a problem is if x = 1 or x = -1



Im hoping someone can provide me with a clear explanation as I am very confused






proof-verification summation induction







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    I am having trouble seeing how this summation equivalence holds true:



    $sum_{i=0}^infty x^i$ = $frac{1}{1-x}$ if |x| < 1



    The only thing I can see where there would be a problem is if x = 1 or x = -1



    Im hoping someone can provide me with a clear explanation as I am very confused






    proof-verification summation induction







    share|cite|improve this question









    $endgroup$















      1












      1








      1


      0



      $begingroup$


      I am having trouble seeing how this summation equivalence holds true:



      $sum_{i=0}^infty x^i$ = $frac{1}{1-x}$ if |x| < 1



      The only thing I can see where there would be a problem is if x = 1 or x = -1



      Im hoping someone can provide me with a clear explanation as I am very confused






      proof-verification summation induction







      share|cite|improve this question









      $endgroup$




      I am having trouble seeing how this summation equivalence holds true:



      $sum_{i=0}^infty x^i$ = $frac{1}{1-x}$ if |x| < 1



      The only thing I can see where there would be a problem is if x = 1 or x = -1



      Im hoping someone can provide me with a clear explanation as I am very confused






      proof-verification summation induction




      proof-verification summation induction






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 5 '18 at 2:11









      ResalexResalex

      61




      61






















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          I actually thought about this earlier and came up with a somewhat heuristic argument. So we know, for a geometric series with ratio $r$,



          $$sum_{n=0}^k r^n = 1 + r + r^2 + r^3 + ... + r^k = frac{r^{k+1}-1}{r-1} = frac{1 - r^{k+1}}{1-r}$$



          (Either of the fractions on the right are valid; which you choose is merely a matter of personal taste. You can see their equivalence by multiplying the top and bottom by $-1$. We will be considering the latter however since it's closer to the formula that most people see.)



          Consider the limiting behavior of this as $k to infty$, i.e. as we accumulate infinitely many terms. What happens? Well, we have...



          $$sum_{n=0}^infty r^n = lim_{k to infty} frac{1 - r^{k+1}}{1-r}$$



          Well, suppose $r = 1$. Well that's dumb, the limit's undefined, so it's really hard to observe anything there. (Of course, we can show that the limit would be infinity by instead considering the limit of the partial sums.)



          Suppose $|r| > 1$. Then clearly, owing to the $r^{k+1}$ term, the fractions increases in magnitude without bound, i.e. diverges. So not much there.



          But what if $|r| < 1$? Then $r^{k+1} to 0$ - it shouldn't be difficult to convince yourself of that much.



          Thus, we say, for $|r|<1$,



          $$sum_{n=0}^infty r^n = frac{1}{1-r}$$



          and for $|r| geq 1$, it diverges.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            Consider the formula for the sum of a finite geometric series:
            $$ sum_{i=0}^n x^i = frac{1-x^n}{1-x}.$$
            Now, if we have that the $|x| < 1$, then when we repeatedly multiply $x$ by itself, it gets smaller. This may give some intuition as to why we can then let $n$ tend to infinity in this formula to give
            $$ sum_{i=1}^infty x^i = frac{1}{1-x},$$ but only when $|x| < 1$, as then $x^n$ tends to 0 as $n$ tends to infinity. We say that 1 is the radius of convergence of the sum - note that $1$ is not included in this radius, so the sum is undefined when $x$ is 1. This should make sense, since $1 + 1 + 1 + cdots $ intuitively doesn't seem to look like it might end up being finite, right?






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              isnt it undefined because when x is 1 you are dividng by 0?
              $endgroup$
              – Resalex
              Dec 5 '18 at 2:23










            • $begingroup$
              Yes exactly - we don't actually divide by 0 though, since we are only asserting that the right hand side equals the left hand side "when $|x| < 1$"
              $endgroup$
              – Stuartg98
              Dec 5 '18 at 2:25












            • $begingroup$
              Does that make sense?
              $endgroup$
              – Stuartg98
              Dec 5 '18 at 2:26






            • 1




              $begingroup$
              ah okay yes it does thank you very much
              $endgroup$
              – Resalex
              Dec 5 '18 at 2:26










            • $begingroup$
              No problem! Glad I could help.
              $endgroup$
              – Stuartg98
              Dec 5 '18 at 2:28











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            2 Answers
            2






            active

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            2 Answers
            2






            active

            oldest

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            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            I actually thought about this earlier and came up with a somewhat heuristic argument. So we know, for a geometric series with ratio $r$,



            $$sum_{n=0}^k r^n = 1 + r + r^2 + r^3 + ... + r^k = frac{r^{k+1}-1}{r-1} = frac{1 - r^{k+1}}{1-r}$$



            (Either of the fractions on the right are valid; which you choose is merely a matter of personal taste. You can see their equivalence by multiplying the top and bottom by $-1$. We will be considering the latter however since it's closer to the formula that most people see.)



            Consider the limiting behavior of this as $k to infty$, i.e. as we accumulate infinitely many terms. What happens? Well, we have...



            $$sum_{n=0}^infty r^n = lim_{k to infty} frac{1 - r^{k+1}}{1-r}$$



            Well, suppose $r = 1$. Well that's dumb, the limit's undefined, so it's really hard to observe anything there. (Of course, we can show that the limit would be infinity by instead considering the limit of the partial sums.)



            Suppose $|r| > 1$. Then clearly, owing to the $r^{k+1}$ term, the fractions increases in magnitude without bound, i.e. diverges. So not much there.



            But what if $|r| < 1$? Then $r^{k+1} to 0$ - it shouldn't be difficult to convince yourself of that much.



            Thus, we say, for $|r|<1$,



            $$sum_{n=0}^infty r^n = frac{1}{1-r}$$



            and for $|r| geq 1$, it diverges.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              I actually thought about this earlier and came up with a somewhat heuristic argument. So we know, for a geometric series with ratio $r$,



              $$sum_{n=0}^k r^n = 1 + r + r^2 + r^3 + ... + r^k = frac{r^{k+1}-1}{r-1} = frac{1 - r^{k+1}}{1-r}$$



              (Either of the fractions on the right are valid; which you choose is merely a matter of personal taste. You can see their equivalence by multiplying the top and bottom by $-1$. We will be considering the latter however since it's closer to the formula that most people see.)



              Consider the limiting behavior of this as $k to infty$, i.e. as we accumulate infinitely many terms. What happens? Well, we have...



              $$sum_{n=0}^infty r^n = lim_{k to infty} frac{1 - r^{k+1}}{1-r}$$



              Well, suppose $r = 1$. Well that's dumb, the limit's undefined, so it's really hard to observe anything there. (Of course, we can show that the limit would be infinity by instead considering the limit of the partial sums.)



              Suppose $|r| > 1$. Then clearly, owing to the $r^{k+1}$ term, the fractions increases in magnitude without bound, i.e. diverges. So not much there.



              But what if $|r| < 1$? Then $r^{k+1} to 0$ - it shouldn't be difficult to convince yourself of that much.



              Thus, we say, for $|r|<1$,



              $$sum_{n=0}^infty r^n = frac{1}{1-r}$$



              and for $|r| geq 1$, it diverges.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                I actually thought about this earlier and came up with a somewhat heuristic argument. So we know, for a geometric series with ratio $r$,



                $$sum_{n=0}^k r^n = 1 + r + r^2 + r^3 + ... + r^k = frac{r^{k+1}-1}{r-1} = frac{1 - r^{k+1}}{1-r}$$



                (Either of the fractions on the right are valid; which you choose is merely a matter of personal taste. You can see their equivalence by multiplying the top and bottom by $-1$. We will be considering the latter however since it's closer to the formula that most people see.)



                Consider the limiting behavior of this as $k to infty$, i.e. as we accumulate infinitely many terms. What happens? Well, we have...



                $$sum_{n=0}^infty r^n = lim_{k to infty} frac{1 - r^{k+1}}{1-r}$$



                Well, suppose $r = 1$. Well that's dumb, the limit's undefined, so it's really hard to observe anything there. (Of course, we can show that the limit would be infinity by instead considering the limit of the partial sums.)



                Suppose $|r| > 1$. Then clearly, owing to the $r^{k+1}$ term, the fractions increases in magnitude without bound, i.e. diverges. So not much there.



                But what if $|r| < 1$? Then $r^{k+1} to 0$ - it shouldn't be difficult to convince yourself of that much.



                Thus, we say, for $|r|<1$,



                $$sum_{n=0}^infty r^n = frac{1}{1-r}$$



                and for $|r| geq 1$, it diverges.






                share|cite|improve this answer









                $endgroup$



                I actually thought about this earlier and came up with a somewhat heuristic argument. So we know, for a geometric series with ratio $r$,



                $$sum_{n=0}^k r^n = 1 + r + r^2 + r^3 + ... + r^k = frac{r^{k+1}-1}{r-1} = frac{1 - r^{k+1}}{1-r}$$



                (Either of the fractions on the right are valid; which you choose is merely a matter of personal taste. You can see their equivalence by multiplying the top and bottom by $-1$. We will be considering the latter however since it's closer to the formula that most people see.)



                Consider the limiting behavior of this as $k to infty$, i.e. as we accumulate infinitely many terms. What happens? Well, we have...



                $$sum_{n=0}^infty r^n = lim_{k to infty} frac{1 - r^{k+1}}{1-r}$$



                Well, suppose $r = 1$. Well that's dumb, the limit's undefined, so it's really hard to observe anything there. (Of course, we can show that the limit would be infinity by instead considering the limit of the partial sums.)



                Suppose $|r| > 1$. Then clearly, owing to the $r^{k+1}$ term, the fractions increases in magnitude without bound, i.e. diverges. So not much there.



                But what if $|r| < 1$? Then $r^{k+1} to 0$ - it shouldn't be difficult to convince yourself of that much.



                Thus, we say, for $|r|<1$,



                $$sum_{n=0}^infty r^n = frac{1}{1-r}$$



                and for $|r| geq 1$, it diverges.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 5 '18 at 2:24









                Eevee TrainerEevee Trainer

                5,7871936




                5,7871936























                    1












                    $begingroup$

                    Consider the formula for the sum of a finite geometric series:
                    $$ sum_{i=0}^n x^i = frac{1-x^n}{1-x}.$$
                    Now, if we have that the $|x| < 1$, then when we repeatedly multiply $x$ by itself, it gets smaller. This may give some intuition as to why we can then let $n$ tend to infinity in this formula to give
                    $$ sum_{i=1}^infty x^i = frac{1}{1-x},$$ but only when $|x| < 1$, as then $x^n$ tends to 0 as $n$ tends to infinity. We say that 1 is the radius of convergence of the sum - note that $1$ is not included in this radius, so the sum is undefined when $x$ is 1. This should make sense, since $1 + 1 + 1 + cdots $ intuitively doesn't seem to look like it might end up being finite, right?






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      isnt it undefined because when x is 1 you are dividng by 0?
                      $endgroup$
                      – Resalex
                      Dec 5 '18 at 2:23










                    • $begingroup$
                      Yes exactly - we don't actually divide by 0 though, since we are only asserting that the right hand side equals the left hand side "when $|x| < 1$"
                      $endgroup$
                      – Stuartg98
                      Dec 5 '18 at 2:25












                    • $begingroup$
                      Does that make sense?
                      $endgroup$
                      – Stuartg98
                      Dec 5 '18 at 2:26






                    • 1




                      $begingroup$
                      ah okay yes it does thank you very much
                      $endgroup$
                      – Resalex
                      Dec 5 '18 at 2:26










                    • $begingroup$
                      No problem! Glad I could help.
                      $endgroup$
                      – Stuartg98
                      Dec 5 '18 at 2:28
















                    1












                    $begingroup$

                    Consider the formula for the sum of a finite geometric series:
                    $$ sum_{i=0}^n x^i = frac{1-x^n}{1-x}.$$
                    Now, if we have that the $|x| < 1$, then when we repeatedly multiply $x$ by itself, it gets smaller. This may give some intuition as to why we can then let $n$ tend to infinity in this formula to give
                    $$ sum_{i=1}^infty x^i = frac{1}{1-x},$$ but only when $|x| < 1$, as then $x^n$ tends to 0 as $n$ tends to infinity. We say that 1 is the radius of convergence of the sum - note that $1$ is not included in this radius, so the sum is undefined when $x$ is 1. This should make sense, since $1 + 1 + 1 + cdots $ intuitively doesn't seem to look like it might end up being finite, right?






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      isnt it undefined because when x is 1 you are dividng by 0?
                      $endgroup$
                      – Resalex
                      Dec 5 '18 at 2:23










                    • $begingroup$
                      Yes exactly - we don't actually divide by 0 though, since we are only asserting that the right hand side equals the left hand side "when $|x| < 1$"
                      $endgroup$
                      – Stuartg98
                      Dec 5 '18 at 2:25












                    • $begingroup$
                      Does that make sense?
                      $endgroup$
                      – Stuartg98
                      Dec 5 '18 at 2:26






                    • 1




                      $begingroup$
                      ah okay yes it does thank you very much
                      $endgroup$
                      – Resalex
                      Dec 5 '18 at 2:26










                    • $begingroup$
                      No problem! Glad I could help.
                      $endgroup$
                      – Stuartg98
                      Dec 5 '18 at 2:28














                    1












                    1








                    1





                    $begingroup$

                    Consider the formula for the sum of a finite geometric series:
                    $$ sum_{i=0}^n x^i = frac{1-x^n}{1-x}.$$
                    Now, if we have that the $|x| < 1$, then when we repeatedly multiply $x$ by itself, it gets smaller. This may give some intuition as to why we can then let $n$ tend to infinity in this formula to give
                    $$ sum_{i=1}^infty x^i = frac{1}{1-x},$$ but only when $|x| < 1$, as then $x^n$ tends to 0 as $n$ tends to infinity. We say that 1 is the radius of convergence of the sum - note that $1$ is not included in this radius, so the sum is undefined when $x$ is 1. This should make sense, since $1 + 1 + 1 + cdots $ intuitively doesn't seem to look like it might end up being finite, right?






                    share|cite|improve this answer









                    $endgroup$



                    Consider the formula for the sum of a finite geometric series:
                    $$ sum_{i=0}^n x^i = frac{1-x^n}{1-x}.$$
                    Now, if we have that the $|x| < 1$, then when we repeatedly multiply $x$ by itself, it gets smaller. This may give some intuition as to why we can then let $n$ tend to infinity in this formula to give
                    $$ sum_{i=1}^infty x^i = frac{1}{1-x},$$ but only when $|x| < 1$, as then $x^n$ tends to 0 as $n$ tends to infinity. We say that 1 is the radius of convergence of the sum - note that $1$ is not included in this radius, so the sum is undefined when $x$ is 1. This should make sense, since $1 + 1 + 1 + cdots $ intuitively doesn't seem to look like it might end up being finite, right?







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 5 '18 at 2:21









                    Stuartg98Stuartg98

                    586




                    586












                    • $begingroup$
                      isnt it undefined because when x is 1 you are dividng by 0?
                      $endgroup$
                      – Resalex
                      Dec 5 '18 at 2:23










                    • $begingroup$
                      Yes exactly - we don't actually divide by 0 though, since we are only asserting that the right hand side equals the left hand side "when $|x| < 1$"
                      $endgroup$
                      – Stuartg98
                      Dec 5 '18 at 2:25












                    • $begingroup$
                      Does that make sense?
                      $endgroup$
                      – Stuartg98
                      Dec 5 '18 at 2:26






                    • 1




                      $begingroup$
                      ah okay yes it does thank you very much
                      $endgroup$
                      – Resalex
                      Dec 5 '18 at 2:26










                    • $begingroup$
                      No problem! Glad I could help.
                      $endgroup$
                      – Stuartg98
                      Dec 5 '18 at 2:28


















                    • $begingroup$
                      isnt it undefined because when x is 1 you are dividng by 0?
                      $endgroup$
                      – Resalex
                      Dec 5 '18 at 2:23










                    • $begingroup$
                      Yes exactly - we don't actually divide by 0 though, since we are only asserting that the right hand side equals the left hand side "when $|x| < 1$"
                      $endgroup$
                      – Stuartg98
                      Dec 5 '18 at 2:25












                    • $begingroup$
                      Does that make sense?
                      $endgroup$
                      – Stuartg98
                      Dec 5 '18 at 2:26






                    • 1




                      $begingroup$
                      ah okay yes it does thank you very much
                      $endgroup$
                      – Resalex
                      Dec 5 '18 at 2:26










                    • $begingroup$
                      No problem! Glad I could help.
                      $endgroup$
                      – Stuartg98
                      Dec 5 '18 at 2:28
















                    $begingroup$
                    isnt it undefined because when x is 1 you are dividng by 0?
                    $endgroup$
                    – Resalex
                    Dec 5 '18 at 2:23




                    $begingroup$
                    isnt it undefined because when x is 1 you are dividng by 0?
                    $endgroup$
                    – Resalex
                    Dec 5 '18 at 2:23












                    $begingroup$
                    Yes exactly - we don't actually divide by 0 though, since we are only asserting that the right hand side equals the left hand side "when $|x| < 1$"
                    $endgroup$
                    – Stuartg98
                    Dec 5 '18 at 2:25






                    $begingroup$
                    Yes exactly - we don't actually divide by 0 though, since we are only asserting that the right hand side equals the left hand side "when $|x| < 1$"
                    $endgroup$
                    – Stuartg98
                    Dec 5 '18 at 2:25














                    $begingroup$
                    Does that make sense?
                    $endgroup$
                    – Stuartg98
                    Dec 5 '18 at 2:26




                    $begingroup$
                    Does that make sense?
                    $endgroup$
                    – Stuartg98
                    Dec 5 '18 at 2:26




                    1




                    1




                    $begingroup$
                    ah okay yes it does thank you very much
                    $endgroup$
                    – Resalex
                    Dec 5 '18 at 2:26




                    $begingroup$
                    ah okay yes it does thank you very much
                    $endgroup$
                    – Resalex
                    Dec 5 '18 at 2:26












                    $begingroup$
                    No problem! Glad I could help.
                    $endgroup$
                    – Stuartg98
                    Dec 5 '18 at 2:28




                    $begingroup$
                    No problem! Glad I could help.
                    $endgroup$
                    – Stuartg98
                    Dec 5 '18 at 2:28


















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