evaluate $\lim_{n\to\infty}\sum_{r=1}^{n-1}\frac{e^{r/n}}{n}$ Each term in the equation given in title tends to zero.
$e^{\frac{r}{n}}$ tends to 1 and the denominator tends to infinity. 
Also, even the greatest numerator in the summation $e^{\frac{n-1}{n}}$ tends to $e$ since $\frac{n-1}{n}=1-(1/n)$ tends to 1 as $n\to\infty$.
But if I apply GP formula (numerator is in GP) I get $e-1$ as the answer which is also given the book.
$$=\frac{e-1}{n(e^{1/n}-1)}=\frac{e-1}{\frac{e^(1/n)-1}{1/n}}$$
And when $n\to\infty$, you get $e-1$.
What is wrong in my first method?
To support my answer, you can also write the expression as $$\int_0^1e^xdx=e-1$$.
 A: hint: Riemann Sum $\to S = \displaystyle \int_{0}^ 1 e^x dx = e-1$. You can continue with your method. In that case, you need to take the limit:
$\displaystyle \lim_{n\to \infty} \dfrac{e^{\frac{1}{n}}-1}{\frac{1}{n}}= \displaystyle \lim_{x\to 0} \dfrac{e^x-1}{x}= \displaystyle \lim_{x\to 0}\dfrac{e^x}{1} = 1$ and get the answer $e-1$. Thus your method works !
A: Your error is that limits do not commute in general (or here $\lim_{n\to\infty}$ and $\sum_{r=0}^\infty$; note that the $\sigma$ cannot be considered a sum because the number of summands is not fixed)
A: Edit: OP in a comment writes that the sum in the title should start at $r=0$ instead of the current $i=1$. With that modification, the calculation becomes correct.
Your method is essentially correct. The sum $x+x^2+\cdots +x^{n-1}$, for $x\ne 1$ is equal to 
$$\frac{x(x^{n-1}-1)}{x-1}.$$ 
In our particular case $x=e^{1/n}$, we get numerator $e-e^{1/n}$ which is a little different from yours. But it is a difference that in the limit makes no difference.
A: To illustrate what's wrong with your first method I take an example: calculate the limit
$$
\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{n}.
$$
Your argument is: all terms goes to zero then... 
Actually there is no then in this case, because each term goes to zero, but their number becomes huge, so that all together they may make the difference. And they do since the sum is $n\frac{1}{n}=1$ all the time when $n\to\infty$.
Conclusion: a growing number of vanishing terms may be non-zero.
