# Studying the convergence/divergence of an infinite sum.

I am studying how to see if a sequence converges or diverges using convergence tests: ratio, comparison, root, integral, limit comparison...

I don't see which convergence test should I apply to this series:

$$\sum_{n\geq 1}\frac{\sqrt{2^n+n^2}}{3^{n-2}}$$

• Use limit-comparison, noting that $2^n+n^2$ is dominated by $2^n$. – Lord Shark the Unknown Sep 25 '17 at 18:08

Use comparison test $$\sum_{n\geq 5}\frac{\sqrt{2^n+n^2}}{3^{n-2}}\leq\sum_{n\geq 5}\frac{\sqrt{2^n+2^n}}{3^{n-2}}$$

• Really I didn't see Lord comment, but I don't feel good! – Nosrati Sep 25 '17 at 18:19

Surely $\sqrt{2^n+n^2}< 2^{n+1}$ for any $n\ge 1$

therefore we have

$\dfrac{\sqrt{2^n+n^2}}{3^{n-2}}<2\cdot 3^2\cdot \dfrac{2^n}{3^n}=18\left(\dfrac{2}{3}\right)^n$

which converges because it's a geometric series with ratio less than $1$

being smaller than a convergent series the given series converges

Hope this helps

Simpler: as it is a series with positive terms, you can use equivalence :

$\sqrt{2^n+n^2}\sim_\infty \bigl(\sqrt 2\mkern2mu\bigr)^n$, so $\;\dfrac{\sqrt{2^n+n^2}}{3^{n-2}}\sim_\infty \dfrac19\biggl(\dfrac{\sqrt 2\mkern2mu}{3}\biggr)^{\!n},\;$ which is a convergent geometric series.