Convergence of $\sum_{n=1}^\infty \frac{n^2}{2^n}$

I have to decide whether the Series $\sum_{n=1}^\infty \frac{n^2}{2^n}$ Converge or Diverge only by using the Comparison test or by $\frac{1}{n^p}$ test.

What I noticed is that it converges by the De'lamber test ($\lim_{n\to \infty}\left|\frac{a_{n+1}}{a_n}\right|$).

But I'm having trouble finding a way proving it with the provided tests.

• Apply the comparison test with $1/1.5^n$ Mar 6 '16 at 18:36

Hint For $n > 16$ we have $$2^n >n^4$$ which can be easily proven by induction.
You may know that for any $q>1$ we have $q^n>n$ for $n\gg 1$. Letting $q<\sqrt 2$ we thus find $\frac{n^2}{2^n}<\left(\frac {q^2}2\right)^n$ for $n\gg 1$. Hence we can compare with the geometric series $\sum c^n$ with $c:=\frac{q^2}{2}<1$.
Using root test for: $\sum_{n=1}^\infty{\frac{n^2}{2^n}}$
$$\lim_{n\to\infty} \left(\frac{n^2}{2^n}\right)^{\frac{1}{n}} = \lim_{n\to\infty}\left(\frac{n^{\frac{2}{n}}}{2^{\frac{n}{n}}}\right) = \frac{1}{2}<1$$ This means that the series is convergent.
Note: $\lim_{n\to\infty} n^{\frac{1}{n}} = 1$