how to find out series is divergent or convergent for $\sum_{n=1}^\infty \frac{2^n}{n^2}$

Question

\begin{equation}\sum_{n=1}^\infty \frac{2^n}{n^2}\end{equation}

The text book says the above series diverges by the n-th term test, but given no procedures how it was done so, could you some enlighten me, thanks.

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edit: it diverges, that was a typo on me.

Answer 1

For $n \geq 4$ we have $2^n \geq n^2$ such that

$\displaystyle\qquad \sum_{n=1}^\infty \frac{2^n}{n^2} \geq \sum_{n=4}^\infty \frac{2^n}{n^2} \geq \sum_{n=4} 1 = +\infty$

making good use of the comparison test.

Answer 2

Hint: You can use the ration test.

$$L = \lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|.$$

The ratio test states that:

if L < 1 then the series converges absolutely;
if L > 1 then the series does not converge;
if L = 1 or the limit fails to exist, 
         then the test is inconclusive, because there exist both convergent 
         and divergent series that satisfy this case.

Answer 3

It diverges. First you can prove by induction that: $2^n > n^3$ for $\forall n \geq 10$. So: $\dfrac{2^n}{n^2} > n$. Thus $\displaystyle \sum_{n=1}^\infty \dfrac{2^n}{n^2} \geq \displaystyle \sum_{n=1}^\infty n = +\infty$. So: $\displaystyle \sum_{n=1}^\infty \dfrac{2^n}{n^2} = +\infty$

Answer 4

A necessary condition for the convergence of a series $\sum_{n=1}^{\infty}a_n$ is that $\displaystyle{\lim_{n\rightarrow\infty}a_n=0}$, but $\displaystyle{\lim_{n\rightarrow\infty}}\frac{2^n}{n^2}=\displaystyle{\lim_{n\rightarrow\infty}}\frac{e^{n\ln 2}}{n^2}=\infty$.

Hence $\displaystyle{\sum_{n=1}^{\infty}{\frac{2^n}{n^2}}}$ diverges.