Evaluate $\displaystyle\sum_{n=1}^{\infty} \left(\dfrac{2}{n}\right)^n$ upto $5$ decimal places.
My attempt: The above series converges as $\left(\frac{2}{n}\right)^n \le \left(\frac{1}{2}\right)^{n}$ for $n\ge 4$. So the sum is bounded by $2+1+\frac{8}{27}+\frac{1}{8} = 3.421\overline{296}$. Since the question asks us to evaluate the sum acurrately upto $5$ decimal places, i think it will be enough to sum only the first $10$ terms of the series, as the $10th$ term will be $\left( \frac{1}{5} \right)^{10} $ which is $1.024\times 10^{-7}$. The sum of first $10$ terms is $3.37058$, but maybe first $10$ terms are not enough, and with more terms it becomes $3.37059$.
I wasn't able to find the exact sum of this series, I don't know if it's even possible to find the exact sum. I converted it into an integral, which resulted in $\sum_{n=1}^{\infty}2^n\left( 1- \int_{1}^{n}\frac{n}{x^{n+1}}dx \right) $, but this didn't help as the limits are dependent on $n$. Any hints on how to solve this problem?
You have$$\sum_{n=1}^7\left(\frac2n\right)^n\approx3.370\,56.$$Since, $n\geqslant 8\implies\frac2n\leqslant\frac14$,$$\sum_{n=8}^\infty\left(\frac2n\right)^n\leqslant\sum_{n=8}^\infty\frac1{4^n}=\frac1{49\,152}\approx0.000\,02.$$Therefore$$3.370\,56<\sum_{n=1}^\infty\left(\frac2n\right)^n<3.370\,59$$and so the answer is $3.370\,5$.