Hi could anyone help me to solve this. Find the radius of convergence of the power series $$\sum^{\infty}_{n=1}(\int^{2n}_{n} \frac{e^t}{t}dt)y^n.$$
To find the radius of convergence, set $a_n = \int^{2n}_{n}\frac{e^t}{t}dt$ and wanted to find $\lim_{n\to\infty}\frac{a_{n+1}}{a_{n}}$. But all my attempt to integrate $\frac{e^t}{t}$ failed. I also tried using $\frac{e^t}{t} = \sum^{\infty}_{k=0}\frac{t^{n-1}}{n!}$ but it became much more complicated.
Hint: If $a_n$ is the coefficient of $y^{n}$ then $\frac 1 {2n} \int_n^{2n} e^{t} dt \leq a_n \leq \frac 1 n \int_n^{2n} e^{t} dt$. Take $n-$th roots and note that $(e^{2n}-e^{n})^{1/n}=e^{2}(1-\frac 1 {e^{n}})^{1/n} \to e^{2}$. By Squeeze Theorem we see that $a_n^{1/n} \to e^{2}$.