Sum of a series

[Bogdan73]

Can someone help me to fin the sum of this series, maybe step by step?

 

[MarkFL]

Hello, and welcome to FMH!

I would look at expressing the summand as follows:

[MATH]\frac{n^2-1}{(n+1)!}-\frac{n}{(n+1)!}=\frac{n-1}{n!}-\frac{n}{(n+1)!}[/MATH]
Now you should be able to show the series is telescoping.

 

[pka]

View attachment 15022 Can someone help me to fin the sum of this series, maybe step by step?

Use the ratio test to prove convergence.
Here is a partial sum \(\displaystyle {S_m} = \sum\limits_{n = 2}^m {\frac{{{n^2} - n - 1}}{{(n + 1)!}}} = \frac{{(m + 2)! - 2{m^2} - 4m}}{{2(m + 2)!}}\)
Using a basic comparison we can see the sequence of partial sums increases and is bounded.

 

[MarkFL]

Okay, now that the partial sum has been given to you, let's see how it can be developed. I would write:

[MATH]S_n=\sum_{k=2}^{n}\left(\frac{k^2-k-1}{(k+1)!}\right)[/MATH]
Using the hint I provided in post #2, we may write:

[MATH]S_n=\sum_{k=2}^{n}\left(\frac{k-1}{k!}-\frac{k}{(k+1)!}\right)[/MATH]
And thus:

[MATH]S_n=\sum_{k=2}^{n}\left(\frac{k-1}{k!}\right)-\sum_{k=2}^{n}\left(\frac{k}{(k+1)!}\right)[/MATH]
Let's re-index the first sum on the right:

[MATH]S_n=\sum_{k=1}^{n-1}\left(\frac{k}{(k+1)!}\right)-\sum_{k=2}^{n}\left(\frac{k}{(k+1)!}\right)[/MATH]
Now, let's pull off the first term of the first sum and the last term of the second sum:

[MATH]S_n=\frac{1}{2}+\sum_{k=2}^{n-1}\left(\frac{k}{(k+1)!}\right)-\sum_{k=2}^{n-1}\left(\frac{k}{(k+1)!}\right)-\frac{n}{(n+1)!}[/MATH]
And since the two sums now add to zero, we're left with:

[MATH]S_n=\frac{1}{2}-\frac{n}{(n+1)!}[/MATH]
Now, can you compute:

[MATH]S_{\infty}=\lim_{n\to\infty}\left(S_n\right)=?[/MATH]