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Determine the radius of convergence of the following power series.

a) $\sum_{n=1}^{\infty}\frac{ x^{6n+2}}{(1+\frac{1}{n})^{n^2}}$

my attempts: by applying the ratio test i got $ \frac {a_n}{a_{n+1}}$ =$\frac{ x^{6n+2}}{(1+\frac{1}{n})^{n^2}}$.$\frac{(1+\frac{1}{n+1})^{(n+1)^2}}{ x^{6n+8}}$

i got $ \frac {a_n}{a_{n+1}}$ = $\frac{e}{x^6}$

now i don't know ...how to find the radius of convergence of given power series.....Pliz help me

thanks in advance

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From $\frac {a_n}{a_{n+1}} \to \frac{e}{x^6}$ we get

$\frac {a_{n+1}}{a_{n}} \to \frac{x^6}{e}$ .

The ratio test shows that the power series converges for $|x|<e^{1/6}$ and diverges for $|x|>e^{1/6}$, hence the radius of convergence is $e^{1/6}$.

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  • $\begingroup$ thanks a lots @Fred $\endgroup$ – user476275 Apr 30 '18 at 5:32
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First of all, typically we apply the Ratio Test as $\displaystyle \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|$, but you have it upside down. Also, don't forget about the limit! Your statement that $$\color{red}{\frac{a_n}{a_{n+1}}=\frac{e}{x^6}}$$ is false; the correct statement is that $$\color{blue}{\lim_{n\to\infty}}{\frac{a_n}{a_{n+1}}=\frac{e}{x^6}}.$$ Other than that, you did a fine job! You just need to flip it over, restore the correct notation — absolute values and limits, and remember that we want the output of the Ratio Test to be less than $1$ to ensure convergence of a series: $$\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\frac{x^6}{e} \implies \text{the series converges when } \frac{x^6}{e}<1.$$ Solving this inequality — and don't forget the absolute value $|x|$ when you take the sixth root! — you'll find the radius of convergence.

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  • $\begingroup$ Thanks a lots @zipirovich $\endgroup$ – user476275 Apr 30 '18 at 5:32

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