I have to calculate a limit that I can't see how as my homework

Asked Dec 25 '19 at 20:11

Active Dec 25 '19 at 20:11

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Alright so this is my first post in here, and it is my last hope for solving my problem.

I searched on the internet, in books, anywhere and I couldn't find a way to solve this.

I have a series which looks like this:

$S_n = a_1 + a_2 + a_3 + .... + a_n$

Where $a_n = {n\over (n+1)!}$

I have to calculate $\lim\limits_{n \to \infty} S_n$

How should I approach this problem? I don't know how to solve it at all.

asked Dec 25 '19 at 20:11

4

Hint. $\frac{n}{(n+1)!}=\frac{(n+1)-1}{(n+1)!}=\frac{1}{n!}-\frac{1}{(n+1)!}$. – Dec 25 '19 at 20:14
Another hint for another approach: $\sum_{n=1}^\infty \frac{n}{(n+1)!} = \sum_{n=0}^\infty \frac{n}{(n+1)!} = \sum_{n=1}^\infty \frac{n-1}{n!}$. – Nicolas Dec 25 '19 at 20:22
I'm gonna try these. If I will not solve it, I will come with a reply. Thanks a lot to both of you. – Octavian Niculescu Dec 25 '19 at 20:46
@Nicolas I suppose that the second sum should be from 1 to infinity too or am I missing something? – Octavian Niculescu Dec 25 '19 at 20:48
@Nicolas I don't really understand what happeneed there. How does n/(n+1)! equal (n-1)/n! – Octavian Niculescu Dec 25 '19 at 20:50
@BernardPan I just finished doing it your way and the limit is 1. Is that right? – Octavian Niculescu Dec 25 '19 at 20:57
@OctavianNiculescu I think so. You can also refer to the similar question referred on this page for detailed solutions. – Dec 25 '19 at 21:09
@BernardPan thanks a lot:) – Octavian Niculescu Dec 25 '19 at 21:19
No, the second sum starts from $0$ because if we let $n = 0$, we have $\frac{n}{(n+1)!} = 0$ so we basically add $0$ to the sum. Then, we let the sum start at $n = 1$ again but replace $n$ with $n-1$ so we essentially still start at $n = 0$. – Nicolas Dec 26 '19 at 18:08

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