Convergence of $a_n=\frac{\sqrt[n]{e^1}+\sqrt[n]{e^2}+\cdots+\sqrt[n]{e^n}}n, b_n=\frac{\sqrt[n]{e^1}+\sqrt[n]{e^2}+\cdots+\sqrt[n]{e^{2n}}}n$ 
Investigate whether the following sequences are convergent:
$$a_n=\left(\frac{\sqrt[n]{e^1}+\sqrt[n]{e^2}+\cdots+\sqrt[n]{e^n}}{n}\right)$$
$$b_n=\left(\frac{\sqrt[n]{e^1}+\sqrt[n]{e^2}+\cdots+\sqrt[n]{e^{2n}}}{n}\right)$$

Any ideas? I guess one can go with induction; I tried it and it resulted in a disaster.
 A: Note that $a_n=\frac{\sqrt[n]{e}}{n} \cdot (1+\sqrt[n]{e}+\sqrt[n]{e}^2+\dotsc+\sqrt[n]{e}^{n-1})=\frac{\sqrt[n]{e}}{n} \cdot \frac{e-1}{\sqrt[n]{e}-1}=(e-1) \cdot \sqrt[n]{e} \cdot \frac{1}{n(\sqrt[n]{e}-1)}$ using the formula for the geometric series.
Now, using $(1+\frac{1}{n})^n \approx e$ for large $n$ this means that
$\sqrt[n]{e}-1 \approx \frac{1}{n}$ and hence the limit should be $e$.
Can you formalize this argument and try a similar one for $b_n$?
A: $$a_n = \frac{1}{n}\sum_{i=1}^{n} e^{i/n}$$
Trivially, a finite sum of finite numbers is a finite number, and then dividing that by another finite number still yields a finite number. Thus, $a_n$ is convergent for all finite $n$. We then note the following:
$$\int_{a}^{b}f(n)dn = \lim_{n \to \infty}\sum_{i=0}^{n}\bigg(\frac{b-a}{n}\bigg)f\bigg(a+\bigg(\frac{b-a}{n}\bigg)i\bigg)$$
Substituting in $b=1$, $a=0$, and $f(n) = e^n$
$$= \lim_{n \to \infty}\sum_{i=1}^{n}\bigg(\frac{1}{n}\bigg)\exp\bigg(\frac{1}{n}\bigg)i$$
$$= \lim_{n \to \infty}\sum_{i=1}^{n}\bigg(\frac{1}{n}\bigg)\exp\bigg(\frac{1}{n}\bigg)i$$
$$= \lim_{n \to \infty}\bigg(\frac{1}{n}\sum_{i=1}^{n}e^{i/n}\bigg)$$
Noting that this is is the limit of $a_n$ as $n$ approaches $\infty$, we find that we have proven that
$$ \lim_{n \to \infty}a_n = \int_{0}^{1}e^ndn$$
$$***$$
Noting that the same finite conditions we found for $a_n$ hold for $b_n$, let us now solve $\lim_{n \to \infty} b_n$:
$$b_n = \frac{1}{n}\sum_{i=1}^{2n} e^{i/n}$$
We now make the substitution $u=2n$
$$= 2\bigg(\frac{1}{u}\sum_{i=1}^{u} e^{2i/u}\bigg)$$
Substituting in $b=2$, $a=0$, and $f(n) = e^{2n}$ into the formula above:
$$\int_{0}^{2}e^{2n}dn = 2\int_{0}^{2}e^udu$$
$$= 2\lim_{n \to \infty}\sum_{i=0}^{u}\bigg(\frac{2}{u}\bigg)\exp\bigg(\frac{2}{u}\bigg)i$$
$$= 4\bigg(\frac{1}{u}\sum_{i=1}^{u} e^{2i/u}\bigg)$$
$$ = 2\lim_{n \to \infty}b_n$$
Therefore, we find we have proved that
$$\lim_{n \to \infty}b_n  = \frac{1}{2}\int_{0}^{2}e^{2n}dn$$
A: you have $ a=\int_{0}^{1} e^x $ and $b=\frac{1}{2}\int_{0}^{2} e^x $
A: $$a_n=\frac1n\biggl(\sum_{k=1}^n \mathrm e^{\tfrac kn}\biggr)$$
is an upper Riemann sum for the function $\mathrm e^x$. Similarly,
$$b_n=\frac1{n}\biggl(\sum_{k=1}^{2n}\mathrm e^{\tfrac kn}\biggr).$$
