Help with finding the n-th partial sum

[Moistwh]

I found values for s1, s2, s3, s4, but can't find the pattern

 

[MarkFL]

I would look at factoring the denominator of the summand and apply partial fraction decomposition to hopefully obtain a telescoping series.

 

[Moistwh]

Thank you very much

 

[MarkFL]

Okay, well to follow through, I would begin by observing:

[MATH]\frac{8}{n^2+12n+35}=4\left(\frac{1}{n+5}-\frac{1}{n+7}\right)[/MATH]
Thus, we may write:

[MATH]S_n=4\left(\sum_{k=1}^n\frac{1}{k+5}-\sum_{k=1}^n\frac{1}{k+7}\right)[/MATH]
Let's now re-index the second sum:

[MATH]S_n=4\left(\sum_{k=1}^n\frac{1}{k+5}-\sum_{k=3}^{n+2}\frac{1}{k+5}\right)[/MATH]
Now, if we pull off the first 2 terms from the first sum and the last 2 terms from the second, we have:

[MATH]S_n=4\left(\frac{1}{6}+\frac{1}{7}+\sum_{k=3}^n\frac{1}{k+5}-\sum_{k=3}^{n}\frac{1}{k+5}-\frac{1}{n+6}-\frac{1}{n+7}\right)[/MATH]
And so what we're left with, since the sums in the middle add to zero, is:

[MATH]S_n=4\left(\frac{1}{6}+\frac{1}{7}-\frac{1}{n+6}-\frac{1}{n+7}\right)[/MATH]
Or:

[MATH]S_n=\frac{2n(13n+85)}{21(n+6)(n+7)}[/MATH]
Hence:

[MATH]S_{\infty}=\lim_{n\to\infty}S_n=\frac{26}{21}[/MATH]