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

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  • $\begingroup$ 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
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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:

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

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  • $\begingroup$ 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
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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 $e-1$.

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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.)

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