# Limit of a sum , what am I doing wrong?

1. Feb 2, 2016

### Steve Turchin

1. The problem statement, all variables and given/known data
$lim_{n \rightarrow \infty}{\frac{1}{n^2} \sum_{k=1}^{n} ke^{\frac{k}{n}}}$

2. Relevant equations

3. The attempt at a solution
$lim_{n \rightarrow \infty}{\frac{1}{n^2} \sum_{k=1}^{n} ke^{\frac{k}{n}}} \\ = lim_{n \rightarrow \infty}{\frac{1}{n^2} (1e^{\frac{1}{n}}+2e^{\frac{2}{n}}+3e^{\frac{3}{n}}+\ldots+ne^{\frac{n}{n}})} \\ = lim_{n \rightarrow \infty}{\frac{e^{\frac{1}{n}}}{n^2} +\frac{2e^{\frac{2}{n}}}{n^2} + \ldots + \frac{e}{n} } = 0$

I know the limit equals to 1 (Wolfram). Isn't the limit of a sum equals to the sum of the limits? What am I doing wrong?

Last edited: Feb 2, 2016
2. Feb 2, 2016

### RUber

It looks like you are saying that since each individual term in the sum goes to zero, the infinite sum of those small terms will also be zero. This is not true.

3. Feb 2, 2016

### Steve Turchin

Thank you for the reply. It's a sum of $n$ terms.

Any tips on how to calculate an exact form of the sum: $\sum_{k=1}^{n} ke^{\frac{k}{n}}$ ?

4. Feb 2, 2016

### RUber

I would recommend starting with the assumption that the exponential term is on the order of 1.
What is
$\sum_{k=1}^n k$?
That should get you started in the right direction.

5. Feb 2, 2016

### geoffrey159

The limit is $\int_0^1 xe^x \ dx$. Check Rieman sums on Wikipedia to see how it works

6. Feb 2, 2016

### Ray Vickson

For fixed $n$ this is just the sum $\sum_{k=1}^n k x^k$, where we substitute $x = e^{1/n}$ at the end. That is a standard sum, and can using Google, for example. However, if you want to get it for yourself you can do it using a bit of calculus. For any $x$ (before the substitution) we have
$$\sum_{k=1}^n k x^k = \sum_k x \frac{d\, x^k}{dx} = x \frac{d}{dx} \sum_{k=1}^n x^k.$$
Use the well-known formula for the sum $\sum_{k=1}^n x^k$, take the derivative of that formula wrt $x$, then multiply by $x$. Finally, substitute $x = e^{1/n}$.