Limit of a geometric series I am trying to practice for a Precalculus exam, but there's this black sheep that I just can't figure out.
$$\lim_{n\to \infty} \dfrac{\sqrt[n]{e}+\sqrt[n]{e^2}+\sqrt[n]{e^3}+...+\sqrt[n]{e^n}}{n}$$
We use the standard sum of a geometric series usually to solve similar limits
$S_n = a_1 \dfrac{q^n-1}{q-1}$. I tried simplifying the series and got $\sqrt[n]{e+e^2+e^3+...+e^n}$. I tried to use the sum formula to end up with $e \dfrac {e^n-1}{e-1}$. But then I get stuck.
 A: You should be careful, $$\sqrt[n]{e} + \sqrt[n]{e^2} + \cdots + \sqrt[n]{e^n} \neq \sqrt[n]{e + e^2 + \cdots + e^n}$$
Instead you should notice that the numerator could be written as $$\sum_{k=1}^n e^{k/n},$$ and this becomes a Riemann sum:
$$\lim_{n\to \infty} \frac1n \sum_{k=1}^n e^{k/n} = \int_0^1 e^x dx$$
A: Your simplification is invalid, but note that the limit is
$$\lim_{n\to\infty} \frac1n \sum_{k=1}^n e^{\frac kn}$$
Wich can be expanded using the geometric sum formula for $e^{\frac1n}$ to obtain:
$$\ldots= \lim_{n\to\infty} \frac{e^{\frac1n}}n \cdot \frac{e - 1}{e^{\frac1n} - 1}= (e-1) \lim_{n\to\infty} \frac{e^{1/n}}{ne^{1/n} - n} = (e-1)\lim_{n\to\infty} \frac{-\frac1{n^2} e^{1/n}}{-\frac1n e^{1/n}+e^{1/n}-1} = e-1$$
A: Let $a=\sqrt[n]{e}$. Then
$$
\begin{align*}
\lim_{n\to\infty}\frac{\sqrt[n]{e}+\ \cdots\ + \sqrt[n]{e^n}}{n}&=\lim_{n\to\infty}\frac{a+a^2+\ \cdots\ + a^n}{n}\\
\text{(summing a geometric series)}\quad &=\lim_{n\to\infty}\frac{a}{n}\cdot \frac{a^{n}-1}{a-1}\\
&=\lim_{n\to\infty}\frac{\sqrt[n]{e}}{n}\cdot \frac{e-1}{\sqrt[n]{e}-1}\\
&=(e-1)\lim_{n\to\infty}\frac{1/n}{1-e^{-1/n}}\\
&=(e-1)\lim_{x\to 0}\frac{x}{1-e^{-x}}\\
\end{align*}
$$
To evaluate $\lim_{x\to 0}\frac{x}{1-e^{-x}}$, use l'Hospital's rule (both the top and bottom tend to 0). We obtain
$$
\lim_{x\to 0}\frac{x}{1-e^{-x}}=\lim_{x\to 0}\frac{1}{e^{-x}}=1.
$$
Thus your limit is $e-1$.
A: It’s not true that
$$\sqrt[n]e+\sqrt[n]{e^2}+\ldots+\sqrt[n]{e^n}=\sqrt[n]{e+e^2+\ldots+e^n}\;.$$
However, if $a=\sqrt[n]e$, then
$$\sqrt[n]e+\sqrt[n]{e^2}+\ldots+\sqrt[n]{e^n}=a+a^2+\ldots+a^n=\frac{a^{n+1}-a}{a-1}\;,$$
so if you want to use the formula for the sum of a geometric series, you should be looking at
$$\lim_{n\to\infty}\frac{e^{1/n}\left((e^{1/n})^n-1\right)}{n(e^{1/n}-1)}=(e-1)\lim_{n\to\infty}\frac{e^{1/n}}{n(e^{1/n}-1)}\;.$$
This can be handled with l’Hospital’s rule. (There are nicer ways to evaluate the original limit, as at least one answer has already pointed out.)
