Determine the radius of convergence of the following power series.?..
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
$\endgroup$2 Answers
$\begingroup$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 $e^{1/6}$.
$\endgroup$ 1 $\begingroup$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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