Prove the Series Converges or Diverges: $\sum_{n=1}^\infty \frac{5+2n}{(1+2n^2)^2}$

Question

$$\sum_{n=1}^\infty \frac{5+2n}{(1+2n^2)^2}$$

How would I even do this to find if this converges or diverges? I tried doing the Telescoping series test but it just got really confusing.

I got as far as this using partial fraction decomposition:

$$\sum_{n=1}^\infty \frac{5}{1+n^2} + \frac{2n-5n^2}{(1+2n^2)^2}$$

$$Sequence: (\frac{5}{2}-\frac{1}{3}),(1-\frac{16}{81}),(\frac{1}{2} -\frac{29}{361}),...$$

I could really use some step by step instructions on how to do this one. Nothing seems to cancel out at all so I felt like I did it wrong. Did I even pick the right test to do it? This was on a quiz I took today so a thorough explanation wouldn't get me any points.

Answer 1

You may observe that, as $n \to \infty$, $$ \frac{5+2n}{(1+2n^2)^2} \sim \frac{1}{2n^3} $$ then use the comparison test.

Answer 2

We have $$\sum_{n\geq1}\frac{5+2n}{\left(1+2n^{2}\right)^{2}}\leq\frac{5}{4}\sum_{n\geq1}\frac{1}{n^{4}}+\frac{1}{2}\sum_{n\geq1}\frac{1}{n^{3}}=\frac{5}{4}\zeta\left(4\right)+\frac{1}{2}\zeta\left(3\right)\approx1.9534. $$

Answer 3

$g(n)=(5+2n)/(1+2n^2)^2=n^{-3}(2+5n^{-1})/(2+n^{-2})^2=n^{-3}f(n)$ where $\lim_{n\to \infty}f(n)=1/2$. So for all sufficiently large $n$ we have $0<g(n)<n^{-3}$. There are many ways to show that $\sum_nn^{-s}$ converges for $s>1$ : e.g. Integral test, or Cauchy condensation test. And so $\sum_ng(n)$ converges by the comparison test.