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.

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    $\begingroup$ Careful, $\sqrt[n]e+\sqrt[n]e^2+\cdots+\sqrt[n]e^n\ne\sqrt[n]{e+e^2+\cdots+e^n}$. $\endgroup$ – Tim Raczkowski Jun 17 '15 at 19:47

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: $$\ldots= \lim_{n\to\infty} \frac{e^{\frac1n}}n \cdot \frac{e - 1}{e^{\frac1n} - 1}= (e-1) \lim_{n\to\infty} \frac{e^{1/n}}{ne^{1/n} - n} = (e-1)\lim_{n\to\infty} \frac{-\frac1{n^2} e^{1/n}}{-\frac1n e^{1/n}+e^{1/n}-1} = e-1$$

  • $\begingroup$ Welp, such a simple overlook by my part. Thanks to everyone. Selecting this as the correct answer since it was the first posted. $\endgroup$ – user1008964 Jun 17 '15 at 20:20
  • $\begingroup$ Your application of l'Hospital's rule has a mistake in the computation of the derivative in the denominator (remember to use the product rule). $\endgroup$ – pre-kidney Jun 18 '15 at 2:01
  • $\begingroup$ @pre-kidney Fixed. Thanks :) $\endgroup$ – AlexR Jun 18 '15 at 8:20

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:

$$\lim_{n\to \infty} \frac1n \sum_{k=1}^n e^{k/n} = \int_0^1 e^x dx$$

  • $\begingroup$ I love it when these limits turn into Riemann sums $\endgroup$ – Simon S Jun 17 '15 at 19:49
  • $\begingroup$ Yeah me too. I got excited when I saw it @SimonS $\endgroup$ – Joel Jun 17 '15 at 19:50
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    $\begingroup$ Although keep in mind this is for a precalculus exam, so Riemann sums would not be allowed... $\endgroup$ – pre-kidney Jun 17 '15 at 19:56
  • $\begingroup$ Without using the Riemann (upper) sum you can use l'Hospital to get the same answer - It's uglier but needs less backing theory :-) $\endgroup$ – AlexR Jun 17 '15 at 20:04
  • $\begingroup$ @pre-kidney, I guess I just saw the Calculus tag. Missed the bit about the precalc exam. Oh well. Still, using l'Hospital's rule would be a bit of calculus... I wonder what the intention of the examiner was. $\endgroup$ – Joel Jun 18 '15 at 19:57

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 $e-1$.


It’s not true that


However, if $a=\sqrt[n]e$, then


so if you want to use the formula for the sum of a geometric series, you should be looking at


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.)


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