How do I find if the series $\sum\frac{2^n}{n^2}$ converges?

Question

How do I find if the series

$$\sum_{n \ge 1}\frac{2^n}{n^2}$$

converges?

I know it diverges but I'm trying to figure out the steps. I tried applying l'hoptital's rule for the divergence test but the result keeps getting bigger... I'm sure there is some simple trick that I'm forgetting but it's driving me nuts. Someone please explain the steps for breaking this one down.

Answer 1

l'hoptital's rule is the way to go.

$$\lim_{n \rightarrow \infty} \frac{2^n}{n^2}=\lim_{n \rightarrow \infty} \frac{\log(2) \cdot 2^{n}}{2 \cdot n} = \lim_{n \rightarrow \infty} \frac{\log(2)^2 \cdot 2^{n}}{2 }$$

So your summands keep getting bigger and bigger.

Answer 2

Hint: What is $$\lim_{n \to \infty} \frac{2^n}{n^2}$$

In order to have convergence, the terms must go to $0$.

Answer 3

Ratio test $$L=\frac{a_{n+1}}{a_n}=\frac{2^{n+1}}{(n+1)^2}\frac{n^2}{2^n}=\frac{2}{(1+\frac{1}{n})^2}$$ the $L>1$, so the series diverges

Answer 4

What about simply the $\;n\,-$ th root test?:

$$\sqrt[n]{\frac{2^n}{n^2}}=\frac2{\left(\sqrt[n]n\right)^2}\xrightarrow[n\to\infty]{}2>1$$

and thus the series diverges.

Answer 5

Note $2^n > n^2$ for $n \geq 5$ by induction. You need to show $2^n > 2n+1$ for the inductive step; to show this, you can use induction again and here it suffices to show $2^n > 2$ which is trivial.

Then $$\sum_{n \geq 1} \frac{2^n}{n^2} > \sum_{n \geq 5} \frac{2^n}{n^2} > \sum_{n \geq 5} 1 \xrightarrow[n \to \infty]\ \infty$$

Asymptotically, $n^2 = o(2^n)$ so not only does the series diverge, but it really diverges.