# A limit calculation using Riemann integral

I was trying to calculate the following limit: $\lim_{n \to \infty}\sum_{k=1}^n \frac {(n+1)^k} {n^{k+1}}$

I think that using some sort of Riemann sum here is the key to manage it, and I got it to the form of: $\lim_{n \to \infty}\sum_{k=1}^n \frac {(n+1)^k} {n^{k+1}} = \lim_{n \to \infty}\sum_{k=1}^n (1+\frac 1 n)^k \ \frac 1 n$

But I can't think of how to continue from here... The way I know Reimann sums is that you have got to have some sort of $\sum_{k=1}^n f(\frac k n) \frac 1 n$ but I cant see such a thing here... Can you give me some hints? :)

Such sum is simply related with a geometric series: $$\frac{1}{n}\sum_{k=1}^{n}\left(1+\frac{1}{n}\right)^k = \frac{n+1}{n}\left[-1+\left(1+\frac{1}{n}\right)^n\right]$$ hence the wanted limit is $\color{red}{\large e-1}$.

• A geometric series! And I go about Riemann sums... I totally should have realized that myself. Thank you very much!! – AsafHaas Mar 24 '17 at 13:20

hint

If you put $a=1+\frac {1}{n}$ then

$$\sum_{k=1}^n a^k=an (a^n-1)$$

may be you will not need Riemann sum.

your limit is $e-1$ since $a^n \to e$.