Confused as to which test to use to determine if series converges or diverges For the question $$\sum^\infty_{n=0}\frac{n^2}{2^n+1}$$I first tried the root test as the denominator was a number to the power of n, but it would result in the numerator having a power to the n so I scrapped that idea. I tried to use the divergence test and then l'hopital's rule as both numerator and denominator went to infinity but I feel like that's over-complicating the question and that there's an easier test for it. Please help.
 A: You can combine the comparison test $\sum \frac{n^2}{2^n+1} < \sum \frac{n^2}{2^n}$ with the ratio test, since
$$
\lim_{n\to\infty} \bigg| \frac{(n+1)^2/2^{n+1}}{n^2/2^n} \bigg| = \lim_{n\to\infty} \frac{n^2+2n+1}{2n^2} = \frac12,
$$
indicating convergence.
You can also use the ratio test directly on the original sum; the limit to evaluate is a bit more complicated, but still very doable.
A: 
We need not apply the ratio test, but rather rely on the comparison test.  All we need is to make use of the binomial theorem and elementary analysis.  To that end we proceed.


First we take $n>3$.  Then, from the binomial theorem we see that 
$$\begin{align}
2^n&=\sum_{k=0}^n \binom{n}{k}\\\\
&\ge \binom{n}{4}\\\\
&=\frac1{24}n(n-1)(n-2)(n-3)\tag1
\end{align}$$

Using $(1)$, we can write for $n\ge 4$
$$\begin{align}
\frac{n^2}{2^n+1}&\le \frac{n^2}{2^n}\\\\
&\le \frac{n^2}{\frac1{24} n(n-1)(n-2)(n-3)}\\\\
&\le \frac{32}{(n-2)(n-3)}\tag2
\end{align}$$

If we now restrict $n$ so that $n\ge 6$, then $(n-2)(n-3)\ge \frac14 n^2$.  Using this estimate in $(2)$ reveals that 
$$\frac{n^2}{2^n}\le \frac{128}{n^2}$$

Inasmuch as the series $\sum_{n=6}^\infty \frac1{n^2}$ converges, then by comparison the series of interest converges also.  And we are done! 
A: IMO ratio test is straight forward:
$$\frac{\frac{(n+1)^2}{2^{n+1}+1}}{\frac{n^2}{2^n+1}}= \frac{(n+1)^2}{n^2}\cdot \frac{2^n+1}{2^{n+1}+1} =\left(1+\frac{1}{n}\right)^2\cdot \frac{1+\frac{1}{2^n} }{2\cdot \left(1+\frac{1}{2^{n+1}} \right)}\stackrel{n \to \infty}{\longrightarrow}\frac{1}{2}$$
