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I think there was a rule in Calculus that mentions this, but I am not sure.

If I need to find $\lim_{n \to \infty} a_n$ and I am only given the nth partial sum: $S_n =\sum_{k=1}^{n} a_k = f(n)$

To find $\lim_{n \to \infty} a_n$ I just have to find $\lim_{n \to \infty} f(n)$ correct?

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    $\begingroup$ Hint: $a_n=f(n)-f(n-1)$ $\endgroup$ – lulu Sep 28 '18 at 14:14
  • $\begingroup$ @lulu so $\lim_{n \to \infty} f(n)-f(n-1)$ I have to find? $\endgroup$ – glockm15 Sep 28 '18 at 14:18
  • $\begingroup$ Yes. $\quad \quad$. $\endgroup$ – lulu Sep 28 '18 at 14:31
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As noticed by lulu in the comment note that

$$S_n-S_{n-1} =\sum_{k=1}^{n} a_k-\sum_{k=1}^{n-1} a_k = a_n\color{red}{+\sum_{k=1}^{n-1} a_k-\sum_{k=1}^{n-1} a_k}=a_n=f(n)-f(n-1)$$


Remark:

  • that is precisely the reason for which $a_n\to 0$ is a necessary condition for the convergence of any series $\sum_{k=1}^{\infty} a_k$, indeed

$$\lim_{n\to \infty}S_n=\sum_{k=1}^{\infty} a_k=L \implies S_n-S_{n-1} =a_n \to 0$$

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  • $\begingroup$ So the limit of my sequence is always 0!? If partial sums exist? $\endgroup$ – glockm15 Sep 28 '18 at 14:25
  • $\begingroup$ @StackUser With reference to the OP we have that $a_n=f(n)-f(n-1)$ therefore $$\lim_{n \to \infty} a_n=\lim_{n \to \infty} f(n)-f(n-1)$$ $\endgroup$ – user Sep 28 '18 at 14:27
  • $\begingroup$ @StackUser The remark given refers to a more general fact about the series, it is not strictly related to your specific example. $\endgroup$ – user Sep 28 '18 at 14:29
  • $\begingroup$ Omm, I am not sure if I get it but, what is happening is that. We are trying to find the last term of $a_n$ which is equivalent to $$\lim_{n \to \infty} a_n$$ and we are doing that by subtracting "the space that is covered" by the partial sums as n goes to infinity to get what the last term will be? $\endgroup$ – glockm15 Sep 28 '18 at 14:31
  • $\begingroup$ @StackUser Do not consider the remark to solve the question, it is another fact we can discuss later. For the OP we need to find $a_n$ and we can use the $S_n=S_{n-1}+a_n\implies a_n=S_n-S_{n-1}$. Here we are using a finite value for $n$. Once we have $a_n$ we can evaluate the limit. $\endgroup$ – user Sep 28 '18 at 14:34

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