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

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

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.

## 3 Answers

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.

• @England The numerator is equivalent to $2n$ ($5$ becomes negligible compared to $2n$ when $n\rightarrow\infty$) The denominator is equivalent to $(2n^2)^2$ – Kitegi Nov 13 '15 at 19:05
• @England You have, as $n$ tends to $\infty$, $$\frac{5+2n}{(1+2n^2)^2}\sim \frac{2n}{(2n^2)^2}=\frac{1}{2n^3}$$ – Olivier Oloa Nov 13 '15 at 19:05
• So as a rule of thumb you can ignore the number being added to the 2n (or a variation of it) as long as it doesn't have an exponent or anything? – England Nov 13 '15 at 19:10
• @England Exactely! – Olivier Oloa Nov 13 '15 at 19:14
• @England It's not just a rule of thumb. There are ways to formalize it and it makes dealing with sequences and series much easier. – Kitegi Nov 13 '15 at 19:54

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

$g(n)=(5+2n)/(1+2n^2)^2=n^{-3}(2+5n^{-1})/(2+n^{-2})^2=n^{-3}f(n)$ 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.