 # The series \sum_{n=1}^\infty n^kr^n coverges when r Holly Guerrero 2021-12-14 Answered
The series $\sum _{n=1}^{\mathrm{\infty }}{n}^{k}{r}^{n}$ coverges when $r\in \left(0,1\right)$ 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 $\sum _{n=1}^{\mathrm{\infty }}{n}^{k}{r}^{n}$. For each series determine the best value of r and decide whether the series converges.
I am stuck on this question $\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{3{n}^{2}+4n+{2}^{-2n}}{{7}^{n+2}+4n+5\sqrt{n}}\right)}^{2}$
I tried using $\frac{{2}^{-2n}}{{7}^{n+2}}$ and ended up with a value of r of $\frac{1}{4\cdot 7}$ but this didn't work.
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Hint:
$4{n}^{2}+4n+{2}^{-2n}\sim 3{n}^{2}\left(n\to +\mathrm{\infty }\right)$
${7}^{n+2}+4n+5\sqrt{n}\sim {7}^{n+2}\left(n\to +\mathrm{\infty }\right)$
the general term of your series is equivalent to
$\frac{9{n}^{4}}{{7}^{2n+4}}$
and has the same nature as
$\sum {n}^{4}{\left(\frac{1}{49}\right)}^{n}$

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Note that
$\underset{n\to \mathrm{\infty }}{lim}\frac{\left(\frac{3{n}^{2}+4n+{2}^{-2n}}{{7}^{n+2}+4n+5\sqrt{n}}{\right)}^{2}}{{n}^{4}{7}^{-2n}}=\underset{n\to \mathrm{\infty }}{lim}\left(\frac{{7}^{n}\left(3{n}^{2}+4n+{2}^{-2n}\right)}{{n}^{2}\left({7}^{n+2}+4n+5\sqrt{n}\right)}{\right)}^{2}$
$=\underset{n\to \mathrm{\infty }}{lim}{\left(\frac{3+4{n}^{-1}+{2}^{-2n}{n}^{-2}}{{7}^{2}+4n{7}^{-n}+5\sqrt{n}{7}^{-n}}\right)}^{2}$
$=\frac{{3}^{2}}{{7}^{4}}$
So, since the series $\sum _{n=1}^{\mathrm{\infty }}\frac{{n}^{4}}{{7}^{2n}}$ converges, your series converges too.

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