# Limit of infinite sequence from partial sum

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?

• Hint: $a_n=f(n)-f(n-1)$ – lulu Sep 28 '18 at 14:14
• @lulu so $\lim_{n \to \infty} f(n)-f(n-1)$ I have to find? – glockm15 Sep 28 '18 at 14:18
• Yes. $\quad \quad$. – lulu Sep 28 '18 at 14:31

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

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

• So the limit of my sequence is always 0!? If partial sums exist? – glockm15 Sep 28 '18 at 14:25
• @StackUser With reference to the OP we have that $a_n=f(n)-f(n-1)$ therefore $$\lim_{n \to \infty} a_n=\lim_{n \to \infty} f(n)-f(n-1)$$ – user Sep 28 '18 at 14:27
• @StackUser The remark given refers to a more general fact about the series, it is not strictly related to your specific example. – user Sep 28 '18 at 14:29
• 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? – glockm15 Sep 28 '18 at 14:31
• @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. – user Sep 28 '18 at 14:34