Limit involving average of exponentials

Question

I am asked to find $$\lim_{n\to\infty} \frac {e^\frac{1}{n} + e^\frac{2}{n} +....+e^\frac{n}{n}}{n}.$$ My attempt is as folllows. By the geometric series formula, $$\frac {e^\frac{1}{n} + e^\frac{2}{n} +....+e^\frac{n}{n}}{n}=\frac{(e-1) e^{1/n}}{n(e^{1/n}-1)}.$$ Therefore, $$\lim_{n\to\infty} \frac {e^\frac{1}{n} + e^\frac{2}{n} +....+e^\frac{n}{n}}{n} = \lim_{n\to\infty} \frac{(e-1) e^{1/n}}{n(e^{1/n}-1)} = \infty.$$ Is this correct?

Answer 1

Another way to do this is to set $1/n = \Delta x$ Then$$ \lim_{n\to\infty} \frac {e^\frac{1}{n} + e^\frac{2}{n} +....+e^\frac{n}{n}}{n}= \lim_{\Delta x\to 0+ }\sum_{k=1}^n{\Delta x e^{k\Delta x}}=\int_0^1{e^x dx} = e-1 $$

Answer 2

$$\frac{\sum_{i=1}^n\exp(i/n)}{n} \le \frac{ne}{n}=e$$

Hence your solution is wrong.

Hint to solve the problem:

Riemann sum.

Answer 3

You are right to claim that by the geometric series formula, $$ \frac{e^{1/n}+\cdots+e^{n/n}}{n}=\frac{\left(e-1\right)e^{1/n}}{n\left(e^{1/n}-1\right)}. $$ However, your next steps are incorrect. We have to proceed carefully since $e^{1/n}-1$ approaches zero as $n$ gets large. Using a Taylor series, $$ n\left(e^{1/n}-1\right)=n\left(\frac{1}{n}+O\left(\frac{1}{n^{2}}\right)\right)=1+O\left(\frac{1}{n}\right). $$ Therefore, $$ \lim_{n\rightarrow\infty}n\left(e^{1/n}-1\right)=1. $$ Putting this all together, $$ \lim_{n\rightarrow\infty}\frac{e^{1/n}+\cdots+e^{n/n}}{n}=e-1. $$

Answer 4

Just added for your curiosity since you already received good answers to the question.

Continuing the Taylor series expansion as in parsiad's answer, we can have very good approximations of the partial sums even for very small values of $n$ $$S_n=\frac{e^{1/n}+\cdots+e^{n/n}}{n}=\frac{\left(e-1\right)e^{1/n}}{n\left(e^{1/n}-1\right)}$$ $$S_n=(e-1)\left(1+\frac{1}{2 n}+\frac{1}{12 n^2}-\frac{1}{720 n^4}+\frac{1}{30240 n^6}-\frac{1}{1209600 n^8} \right)+O\left(\frac{1}{n^{10}}\right)$$ $$\left( \begin{array}{ccc} n & \text{exact} & \text{approximation} \\ 1 & 2.7182818284590452354 & \color{red}{2.718281}7934724389398 \\ 2 & 2.1835015495795866911 & \color{red}{2.1835015495}447755709 \\ 3 & 2.0205427648666035402 & \color{red}{2.02054276486}59977459 \\ 4 & 1.9420071331148973837 & \color{red}{1.9420071331148}632274 \\ 5 & 1.8958338026286923933 & \color{red}{1.8958338026286}887237 \\ 6 & 1.8654476448430682793 & \color{red}{1.86544764484306}76864 \\ 7 & 1.8439374993059529344 & \color{red}{1.843937499305952}8075 \\ 8 & 1.8279112064429921886 & \color{red}{1.8279112064429921}552 \\ 9 & 1.8155093460652278961 & \color{red}{1.8155093460652278}858 \\ 10 & 1.8056275828122667028 & \color{red}{1.805627582812266}6992 \end{array} \right)$$