Try this problem from IIT JAM 2017 exam (Problem 48) and know how to determine radius of convergence of a power series.
Find the radius of convergence of the power series
$$
\sum_{n=1}^{\infty} \frac{(-4)^{n}}{n(n+1)}(x+2)^{2 n}
$$
Real Analysis
Series of Functions
Power Series
Answer: $\frac12$
IIT JAM 2016 , Problem 48
Real Analysis : Robert G. Bartle
Given, the power series is $\sum_{n=1}^{\infty} \frac{(-4)^{n}}{n(n+1)}(x+2)^{2 n}$.
Let us put $2n=m$ to get the standard form of a power series.
We get,
$\sum_{m=2}^{\infty} \frac{(-4)^{\frac m2}}{\frac m2(\frac m2+1)}(x+2)^{ m}$.
Now let us make the transformation $z=x+2$ to get a power series about 0 :
We have,
$\sum_{m=2}^{\infty} \frac{(-4)^{\frac m2}}{\frac m2(\frac m2+1)}(z)^{ m}$
Compairing with $ \sum_{m=2}^{\infty} a_m (z)^m$
we get,
$a_m= \frac{(-4)^{\frac m2}}{\frac m2(\frac m2+1)} $
Now we have to test the convergence of the series.
Can you apply Ratio Test to check the convergence of the series.
Ratio Test : Let $\sum_{n=0}^{\infty} a_{n} x^{n}$ be a power series and let $\lim \left|\frac{a_{n+1}}{a_{n}}\right|=\mu .$ Then
$\begin{aligned}\left|\frac{a_{m+1}}{a_m}\right| &=\left| \frac{4^{\frac m2}\cdot 2\cdot4}{(m+1)(m+3)} \times \frac{m(m+2)}{4^{\frac m2} \cdot 4}\right| \\&=\left| \quad \frac{2\left(1+\frac{2}{m}\right)}{\left(1+\frac{1}{m}\right)(1+\frac 3m)}\right|\end{aligned}$
Now
$\lim \left|\frac{a_{m+1}}{a_{m}}\right|=2 \in (0,\infty)$
Then, The given power series is absolutely convergent i.e., convergent $\forall x$ such that $|x+2|<\frac 12$
Then the answer is $\frac 12$
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