Series convergence geometric series

Question

The series $\sum_{n=1}^\infty n^k r^n$ converges when r ∈(0,1) and diverges when r>1. This is true regardless of the value of the constant k. When r=1 the series is a p-series. It converges if k<-1 and diverges otherwise Each of the series below can be compared to a series of the form $\sum_{n=1}^\infty n^kr^n$. For each series determine the best value of r and decide whether the series converges.

I am stuck on this question $\sum_{n=1}^\infty (\frac{3n^2+4n+2^{-2n}}{7^{n+2}+4n+5\sqrt{n}})^2$.

I tried using $\frac {2^{-2n}}{7^{n+2}}$ and ended up with a value of r of $\frac{1}{4*7}$ but this didn't work.

Answer 1

hint

$$3n^2+4n+2^{-2n}\sim 3n^2 \;\;(n\to +\infty)$$

$$7^{n+2}+4n+5\sqrt{n} \sim 7^{n+2} \;\;(n\to +\infty)$$

the general term of your series is equivalent to

$$\frac{9n^4}{7^{2n+4}}$$ and has the same nature as

$$\sum n^4(\frac{1}{49})^n$$

Answer 2

Note that\begin{align}\lim_{n\to\infty}\frac{\left(\frac{3n^2+4n+2^{-2n}}{7^{n+2}+4n+5\sqrt n}\right)^2}{n^47^{-2n}}&=\lim_{n\to\infty}\left(\frac{7^n(3n^2+4n+2^{-2n})}{n^2(7^{n+2}+4n+5\sqrt n)}\right)^2\\&=\lim_{n\to\infty}\left(\frac{3+4n^{-1}+2^{-2n}n^{-2}}{7^2+4n7^{-n}+5\sqrt n7^{-n}}\right)^2\\&=\frac{3^2}{7^4}.\end{align}So, since the series $\displaystyle\sum_{n=1}^\infty\frac{n^4}{7^{2n}}$ converges, your series converges too.