What is the value of the convergent series $ sum_{n=1}^{infty} (e^{-4n}-e^{-4(n-1)})$? The...

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What is the value of the convergent series $ sum_{n=1}^{infty} (e^{-4n}-e^{-4(n-1)})$?



The Next CEO of Stack OverflowConvergent/divergent series $sum_{n=2}^{infty}(nsqrt{n}-sqrt{n^3-1})$Does $sum_{n = 1}^inftyfrac{ln^2(n)}{sqrt{n}(8n + 9sqrt{n})}$ Converge?Find values of $p$ for which the series is convergent.The Series( $sum_{1}^{+ infty}frac{1}{n! + n}$ convergence or divergence?$sum_{n= 0}^{infty}a_n$ converges, what other series must then also converge?Trying to study to convergence of the series $sum_{n=1}^{infty} frac{ (n!)^2 4^n }{(2n)!}$For what P would the series $sum_{n=2}^infty frac{1}{n^pln(n)}$ converge?Determining whether the series: $sum_{n=1}^{infty} tanleft(frac{1}{n}right) $ converges$sum_limits{n=1}^{infty}(1-cos(frac{pi}{n}))$ convergence proofConvergence of the series $sum_{n=1}^{infty} a^{ln(n)}$












0












$begingroup$


Consider the series
$$
sum_{n=1}^{infty} (e^{-4n}-e^{-4(n-1)}).
$$
I know the series must converge but, what test should I employ to find the value of the series?



Attempt:
I have tried using the integral test, which gives a value of 3/2, but it is incorrect.



The following is the original problem:




enter image description here







calculus sequences-and-series limits







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    This is a telescoping series. Try writing out the first few terms to see the cancellations. Alternatively you can just factor out $e^{-4n}$ and you get a standard geometric sum.
    $endgroup$
    – Michael
    Mar 16 at 19:02












  • $begingroup$
    I assume you know how to evaluate $Asum_{n=1}^{infty} r^n$.
    $endgroup$
    – Michael
    Mar 16 at 19:16










  • $begingroup$
    Assuming the terms of the series are monotonic, the only thing you can get out of the integral test is whether the sum converges. You cannot compute the value of the series this way.
    $endgroup$
    – Wojowu
    Mar 16 at 19:39
















0












$begingroup$


Consider the series
$$
sum_{n=1}^{infty} (e^{-4n}-e^{-4(n-1)}).
$$
I know the series must converge but, what test should I employ to find the value of the series?



Attempt:
I have tried using the integral test, which gives a value of 3/2, but it is incorrect.



The following is the original problem:




enter image description here







calculus sequences-and-series limits







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    This is a telescoping series. Try writing out the first few terms to see the cancellations. Alternatively you can just factor out $e^{-4n}$ and you get a standard geometric sum.
    $endgroup$
    – Michael
    Mar 16 at 19:02












  • $begingroup$
    I assume you know how to evaluate $Asum_{n=1}^{infty} r^n$.
    $endgroup$
    – Michael
    Mar 16 at 19:16










  • $begingroup$
    Assuming the terms of the series are monotonic, the only thing you can get out of the integral test is whether the sum converges. You cannot compute the value of the series this way.
    $endgroup$
    – Wojowu
    Mar 16 at 19:39














0












0








0





$begingroup$


Consider the series
$$
sum_{n=1}^{infty} (e^{-4n}-e^{-4(n-1)}).
$$
I know the series must converge but, what test should I employ to find the value of the series?



Attempt:
I have tried using the integral test, which gives a value of 3/2, but it is incorrect.



The following is the original problem:




enter image description here







calculus sequences-and-series limits







share|cite|improve this question











$endgroup$




Consider the series
$$
sum_{n=1}^{infty} (e^{-4n}-e^{-4(n-1)}).
$$
I know the series must converge but, what test should I employ to find the value of the series?



Attempt:
I have tried using the integral test, which gives a value of 3/2, but it is incorrect.



The following is the original problem:




enter image description here







calculus sequences-and-series limits




calculus sequences-and-series limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 19 at 16:21









Jack

27.6k1782203




27.6k1782203










asked Mar 16 at 18:55









MCCMCC

316




316








  • 1




    $begingroup$
    This is a telescoping series. Try writing out the first few terms to see the cancellations. Alternatively you can just factor out $e^{-4n}$ and you get a standard geometric sum.
    $endgroup$
    – Michael
    Mar 16 at 19:02












  • $begingroup$
    I assume you know how to evaluate $Asum_{n=1}^{infty} r^n$.
    $endgroup$
    – Michael
    Mar 16 at 19:16










  • $begingroup$
    Assuming the terms of the series are monotonic, the only thing you can get out of the integral test is whether the sum converges. You cannot compute the value of the series this way.
    $endgroup$
    – Wojowu
    Mar 16 at 19:39














  • 1




    $begingroup$
    This is a telescoping series. Try writing out the first few terms to see the cancellations. Alternatively you can just factor out $e^{-4n}$ and you get a standard geometric sum.
    $endgroup$
    – Michael
    Mar 16 at 19:02












  • $begingroup$
    I assume you know how to evaluate $Asum_{n=1}^{infty} r^n$.
    $endgroup$
    – Michael
    Mar 16 at 19:16










  • $begingroup$
    Assuming the terms of the series are monotonic, the only thing you can get out of the integral test is whether the sum converges. You cannot compute the value of the series this way.
    $endgroup$
    – Wojowu
    Mar 16 at 19:39








1




1




$begingroup$
This is a telescoping series. Try writing out the first few terms to see the cancellations. Alternatively you can just factor out $e^{-4n}$ and you get a standard geometric sum.
$endgroup$
– Michael
Mar 16 at 19:02






$begingroup$
This is a telescoping series. Try writing out the first few terms to see the cancellations. Alternatively you can just factor out $e^{-4n}$ and you get a standard geometric sum.
$endgroup$
– Michael
Mar 16 at 19:02














$begingroup$
I assume you know how to evaluate $Asum_{n=1}^{infty} r^n$.
$endgroup$
– Michael
Mar 16 at 19:16




$begingroup$
I assume you know how to evaluate $Asum_{n=1}^{infty} r^n$.
$endgroup$
– Michael
Mar 16 at 19:16












$begingroup$
Assuming the terms of the series are monotonic, the only thing you can get out of the integral test is whether the sum converges. You cannot compute the value of the series this way.
$endgroup$
– Wojowu
Mar 16 at 19:39




$begingroup$
Assuming the terms of the series are monotonic, the only thing you can get out of the integral test is whether the sum converges. You cannot compute the value of the series this way.
$endgroup$
– Wojowu
Mar 16 at 19:39










3 Answers
3






active

oldest

votes


















2












$begingroup$

This series converges. Here is the reason:$$sum_{n=1}^{infty} e^{-4n}-e^{-4(n+1)}{=sum_{n=1}^{infty} e^{-4n}-e^{-4n}e^{-4}\=(1-e^{-4})sum_{n=1}^{infty}e^{-4n}\=(1-e^{-4}){e^{-4}over 1-e^{-4}}\=e^{-4}}$$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    The partial sums are $e^{-4}-e^{-4(n+1)}$ by induction, making the limit $e^{-4}$.






    share|cite|improve this answer









    $endgroup$









    • 3




      $begingroup$
      Exploiting the telescoping series is a much more expedient way forward.
      $endgroup$
      – Mark Viola
      Mar 16 at 20:54



















    0












    $begingroup$

    Split the sum$$begin{align*}sumlimits_{ngeq1}left[e^{-4n}-e^{-4(n-1)}right] & =sumlimits_{ngeq1}e^{-4n}-sumlimits_{ngeq1}e^{-4(n-1)}\ & =e^{-4}sumlimits_{ngeq0}e^{-4n}-sumlimits_{ngeq0}e^{-4n}\ & =left(e^{-4}-1right)sumlimits_{ngeq0}e^{-4n}end{align*}$$The infinite sum converges by the infinite geometric sequence which converges for $|x|<1$.$$sumlimits_{ngeq0}x^n=frac 1{1-x}$$It can be shown that $e^{-4}<1$, meaning the sum converges.






    share|cite|improve this answer









    $endgroup$














      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      This series converges. Here is the reason:$$sum_{n=1}^{infty} e^{-4n}-e^{-4(n+1)}{=sum_{n=1}^{infty} e^{-4n}-e^{-4n}e^{-4}\=(1-e^{-4})sum_{n=1}^{infty}e^{-4n}\=(1-e^{-4}){e^{-4}over 1-e^{-4}}\=e^{-4}}$$






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        This series converges. Here is the reason:$$sum_{n=1}^{infty} e^{-4n}-e^{-4(n+1)}{=sum_{n=1}^{infty} e^{-4n}-e^{-4n}e^{-4}\=(1-e^{-4})sum_{n=1}^{infty}e^{-4n}\=(1-e^{-4}){e^{-4}over 1-e^{-4}}\=e^{-4}}$$






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          This series converges. Here is the reason:$$sum_{n=1}^{infty} e^{-4n}-e^{-4(n+1)}{=sum_{n=1}^{infty} e^{-4n}-e^{-4n}e^{-4}\=(1-e^{-4})sum_{n=1}^{infty}e^{-4n}\=(1-e^{-4}){e^{-4}over 1-e^{-4}}\=e^{-4}}$$






          share|cite|improve this answer









          $endgroup$



          This series converges. Here is the reason:$$sum_{n=1}^{infty} e^{-4n}-e^{-4(n+1)}{=sum_{n=1}^{infty} e^{-4n}-e^{-4n}e^{-4}\=(1-e^{-4})sum_{n=1}^{infty}e^{-4n}\=(1-e^{-4}){e^{-4}over 1-e^{-4}}\=e^{-4}}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Mar 16 at 19:31









          Mostafa AyazMostafa Ayaz

          18.1k31040




          18.1k31040























              1












              $begingroup$

              The partial sums are $e^{-4}-e^{-4(n+1)}$ by induction, making the limit $e^{-4}$.






              share|cite|improve this answer









              $endgroup$









              • 3




                $begingroup$
                Exploiting the telescoping series is a much more expedient way forward.
                $endgroup$
                – Mark Viola
                Mar 16 at 20:54
















              1












              $begingroup$

              The partial sums are $e^{-4}-e^{-4(n+1)}$ by induction, making the limit $e^{-4}$.






              share|cite|improve this answer









              $endgroup$









              • 3




                $begingroup$
                Exploiting the telescoping series is a much more expedient way forward.
                $endgroup$
                – Mark Viola
                Mar 16 at 20:54














              1












              1








              1





              $begingroup$

              The partial sums are $e^{-4}-e^{-4(n+1)}$ by induction, making the limit $e^{-4}$.






              share|cite|improve this answer









              $endgroup$



              The partial sums are $e^{-4}-e^{-4(n+1)}$ by induction, making the limit $e^{-4}$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Mar 16 at 19:40









              J.G.J.G.

              32.3k23250




              32.3k23250








              • 3




                $begingroup$
                Exploiting the telescoping series is a much more expedient way forward.
                $endgroup$
                – Mark Viola
                Mar 16 at 20:54














              • 3




                $begingroup$
                Exploiting the telescoping series is a much more expedient way forward.
                $endgroup$
                – Mark Viola
                Mar 16 at 20:54








              3




              3




              $begingroup$
              Exploiting the telescoping series is a much more expedient way forward.
              $endgroup$
              – Mark Viola
              Mar 16 at 20:54




              $begingroup$
              Exploiting the telescoping series is a much more expedient way forward.
              $endgroup$
              – Mark Viola
              Mar 16 at 20:54











              0












              $begingroup$

              Split the sum$$begin{align*}sumlimits_{ngeq1}left[e^{-4n}-e^{-4(n-1)}right] & =sumlimits_{ngeq1}e^{-4n}-sumlimits_{ngeq1}e^{-4(n-1)}\ & =e^{-4}sumlimits_{ngeq0}e^{-4n}-sumlimits_{ngeq0}e^{-4n}\ & =left(e^{-4}-1right)sumlimits_{ngeq0}e^{-4n}end{align*}$$The infinite sum converges by the infinite geometric sequence which converges for $|x|<1$.$$sumlimits_{ngeq0}x^n=frac 1{1-x}$$It can be shown that $e^{-4}<1$, meaning the sum converges.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Split the sum$$begin{align*}sumlimits_{ngeq1}left[e^{-4n}-e^{-4(n-1)}right] & =sumlimits_{ngeq1}e^{-4n}-sumlimits_{ngeq1}e^{-4(n-1)}\ & =e^{-4}sumlimits_{ngeq0}e^{-4n}-sumlimits_{ngeq0}e^{-4n}\ & =left(e^{-4}-1right)sumlimits_{ngeq0}e^{-4n}end{align*}$$The infinite sum converges by the infinite geometric sequence which converges for $|x|<1$.$$sumlimits_{ngeq0}x^n=frac 1{1-x}$$It can be shown that $e^{-4}<1$, meaning the sum converges.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Split the sum$$begin{align*}sumlimits_{ngeq1}left[e^{-4n}-e^{-4(n-1)}right] & =sumlimits_{ngeq1}e^{-4n}-sumlimits_{ngeq1}e^{-4(n-1)}\ & =e^{-4}sumlimits_{ngeq0}e^{-4n}-sumlimits_{ngeq0}e^{-4n}\ & =left(e^{-4}-1right)sumlimits_{ngeq0}e^{-4n}end{align*}$$The infinite sum converges by the infinite geometric sequence which converges for $|x|<1$.$$sumlimits_{ngeq0}x^n=frac 1{1-x}$$It can be shown that $e^{-4}<1$, meaning the sum converges.






                  share|cite|improve this answer









                  $endgroup$



                  Split the sum$$begin{align*}sumlimits_{ngeq1}left[e^{-4n}-e^{-4(n-1)}right] & =sumlimits_{ngeq1}e^{-4n}-sumlimits_{ngeq1}e^{-4(n-1)}\ & =e^{-4}sumlimits_{ngeq0}e^{-4n}-sumlimits_{ngeq0}e^{-4n}\ & =left(e^{-4}-1right)sumlimits_{ngeq0}e^{-4n}end{align*}$$The infinite sum converges by the infinite geometric sequence which converges for $|x|<1$.$$sumlimits_{ngeq0}x^n=frac 1{1-x}$$It can be shown that $e^{-4}<1$, meaning the sum converges.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Mar 16 at 19:40









                  Frank W.Frank W.

                  3,7351321




                  3,7351321






























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