Mrthdrb

For the question $$sum^infty_n=0fracn^22^n+1$$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.

You can combine the comparison test $sum fracn^22^n+1 < sum fracn^22^n$ with the ratio test, since
$$
lim_ntoinfty bigg| frac(n+1)^2/2^n+1n^2/2^n bigg| = lim_ntoinfty fracn^2+2n+12n^2 = frac12,
$$
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.

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

$$beginalign
2^n&=sum_k=0^n binomnk\\
&ge binomn4\\
&=frac124n(n-1)(n-2)(n-3)tag1
endalign$$

Using $(1)$, we can write for $nge 4$

$$beginalign
fracn^22^n+1&le fracn^22^n\\
&le fracn^2frac124 n(n-1)(n-2)(n-3)\\
&le frac32(n-2)(n-3)tag2
endalign$$

If we now restrict $n$ so that $nge 6$, then $(n-2)(n-3)ge frac14 n^2$. Using this estimate in $(2)$ reveals that

$$fracn^22^nle frac128n^2$$

Inasmuch as the series $sum_n=6^infty frac1n^2$ converges, then by comparison the series of interest converges also. And we are done!

For each $ninmathbb Z^+$, $dfracn^22^n+1leqslantdfracn^22^n$. On the other hand, if $n$ is large enough, then $dfracn^22^nleqslantleft(dfrac23right)^n$ because this means that $n^2leqslantleft(dfrac43right)^n$ and an exponential function with a base greater than $1$ always growths faster than a polynomial function. So, since $displaystylesum_n=0^inftyleft(dfrac23right)^n$ converges, your series converges too.

IMO ratio test is straight forward:
$$fracfrac(n+1)^22^n+1+1fracn^22^n+1= frac(n+1)^2n^2cdot frac2^n+12^n+1+1 =left(1+frac1nright)^2cdot frac1+frac12^n 2cdot left(1+frac12^n+1 right)stackreln to inftylongrightarrowfrac12$$