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

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.

• Did you try to move to new log base? en.wikipedia.org/wiki/Logarithm#Change_of_base – stiv Sep 17 '16 at 19:51
• Well, a series can only converge if the limit of the summands is 0. 2^n is exponential and n^2 is geometric so obviously the limits diverges. – fleablood Sep 17 '16 at 19:55
• Are you asking about the series (sum) or the sequence? – fleablood Sep 17 '16 at 20:01
• The ratio test is simple and clear, why not use it – codetalker Sep 17 '16 at 20:02
• L'hopitals yields $\ln 2*2^n/2n$ Apply it twice you get $\ln^2 2^n/2$ which diverges. – fleablood Sep 17 '16 at 20:03

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.

• Do you mean $\log (2)^2$ instead of $\log(n)^2$? – MathMajor Sep 17 '16 at 19:53
• Yes, sorry. I will edit it. – Kaligule Sep 17 '16 at 19:55
• This is exactly where I'm stuck. How did you get from log(2)*2^n/2n to log(2)^2*2^n/2 ? I get a much longer result when I differentiate log(2)*2^n/2n – Tatiana Frank Sep 17 '16 at 20:06
• i think I wasnt treating log(2) as a constant and that's why... – Tatiana Frank Sep 17 '16 at 20:07
• are you using the quotient rule when you derive the second time? – Tatiana Frank Sep 17 '16 at 20:09

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

• Better then my answer, because the ratio test is more universal and it is easier to use. – Kaligule Sep 17 '16 at 19:57

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

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

• Yes I stated in the question description, I tried applying l'hopital's rule for the divergence test and the answer keeps expanding. So your comment is not helpful if you read the description below my question. – Tatiana Frank Sep 17 '16 at 19:54
• @TatianaFrank Applying L'Hopital's rule twice you get $\lim_{n \to \infty} \frac{\log (2)^2 2^n}{2}$ – MathMajor Sep 17 '16 at 19:55
• Your title is asking about the series (sum) not the sequence. The series can not converge if the terms are positive but do not converge to zero. So asking about tne limit of the terms is reasonable. If that was your question, you should have asked that and not about the series. – fleablood Sep 17 '16 at 19:59

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.

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.