$\sum_{n=1}^{\infty}\frac{4n^2+26n+34}{n^4+10n^3+35n^2+50n+24}$ convergent?

Question

serie is convergent?

$$\sum_{n=1}^{\infty}\frac{4n^2+26n+34}{n^4+10n^3+35n^2+50n+24}$$

Factor: $$\frac{4n^2+26n+34}{(n+1)(n+2)(n+3)(n+4)}$$

What test can i Use?

Answer 1

Simply note that

$$\frac{4n^2+26n+34}{n^4+10n^3+35n^2+50n+24}\sim \frac4{n^2}$$

then refer to limit comparison test with $\sum \frac1{n^2}$ which converges.

Answer 2

$$ \frac{4 n^2+26 n+34}{(n+1) (n+2) (n+3) (n+4)} = \frac{1}{n+2}-\frac{4}{n+3}+\frac{1}{n+4}+\frac{2}{n+1} $$

Then

$$ \sum_{n=1}^{\infty}\frac{4 n^2+26 n+34}{(n+1) (n+2) (n+3) (n+4)} = \lim_{n\to\infty}\left(2\sum_{k=2}^{n}\frac{1}{k}+\sum_{k=3}^{n}\frac{1}{k}-4\sum_{k=4}^{n}\frac{1}{k}+\sum_{k=5}^{n}\frac{1}{k}\right)=2\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{3}+\frac{1}{4}\right)-4\frac{1}{4} = \frac{7}{4} $$