Radius of Convergence of a Power series | IIT JAM 2016

Find the radius of convergence of the power series
$$
\sum_{n=1}^{\infty} \frac{(-4)^{n}}{n(n+1)}(x+2)^{2 n}
$$

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.

$\begin{aligned}\left|\frac{a_{m+1}}{a_m}\right| &=\left| \frac{4^{\frac m2}\cdot 2\cdot4}{(m+1)(m+3)} \times \frac{m(m+2)}{4^{\frac m2} \cdot 4}\right| \\&=\left| \quad \frac{2\left(1+\frac{2}{m}\right)}{\left(1+\frac{1}{m}\right)(1+\frac 3m)}\right|\end{aligned}$

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 $\frac 12$