Pointwise and uniform convergence of ${\sum_{n=1}^{+\infty}\left (1+\frac{1}{2n}\right )^{4n^2}x^n}$

Question

I want to check the pointwise and uniform convergence of the below power series.

• $\displaystyle{\sum_{n=1}^{+\infty}\left (1+\frac{1}{2n}\right )^{4n^2}x^n}$

We have that \begin{equation*}\sqrt[n]{\left (1+\frac{1}{2n}\right )^{4n^2}}=\left (1+\frac{1}{2n}\right )^{4n}=\left [\left (1+\frac{1}{2n}\right )^{2n} \right ]^2\to e^2=\frac{1}{R} \Rightarrow R=\frac{1}{e^2}\end{equation*} So we have pointwise convergence for $|x|<\frac{1}{e^2}$.

We check also the boundaries $x=-\frac{1}{e^2}$ and $x=\frac{1}{e^2}$.

At $x=-\frac{1}{e^2}$ we have $\displaystyle{\sum_{n=1}^{+\infty}\left (1+\frac{1}{2n}\right )^{4n^2}\left (-\frac{1}{e^2}\right )^n}$.

At $x=\frac{1}{e^2}$ we have $\displaystyle{\sum_{n=1}^{+\infty}\left (1+\frac{1}{2n}\right )^{4n^2}\left (\frac{1}{e^2}\right )^n}$.

How do we check these series?

Answer 1

You have $$\ln \left[\left( 1 + \frac{1}{2n}\right)^{4n^2} \left( \frac{1}{e^2}\right)^n \right] = 4n^2 \ln \left( 1 + \frac{1}{2n}\right) - 2n = 4n^2 \left( \frac{1}{2n} - \frac{1}{8n^2} + o\left( \frac{1}{n^2}\right)\right) - 2n$$ $$= -\frac{1}{2} + o \left( 1\right)$$

So $$\lim_{n \rightarrow +\infty} \left( 1 + \frac{1}{2n}\right)^{4n^2} \left( \frac{1}{e^2}\right)^n = \frac{1}{\sqrt{e}}$$

so both $$\sum \left( 1 + \frac{1}{2n}\right)^{4n^2} \left( \frac{-1}{e^2}\right)^n \quad \text{ and } \quad \sum \left( 1 + \frac{1}{2n}\right)^{4n^2} \left( \frac{1}{e^2}\right)^n$$ diverge.