Find $\lim_{n \to \infty} \sum^n_{k=1} {{(n+1)^k} \over {n^{k+1}}}$ using Riemann sums

Question

Find$$\lim_{n \to \infty} \sum^n_{k=1} {{(n+1)^k} \over {n^{k+1}}}$$using Riemann sums.

I have: $$\lim_{n \to \infty} \sum^n_{k=1} {{(n+1)^k} \over {n^{k+1}}} = \lim_{n \to \infty} \sum^n_{k=1} (1 + {1 \over n})^k {1 \over n}$$

Now I need to define a function for using Riemann sums, but I don't really see what it should be.

Answer 1

$$\begin{eqnarray*}\frac{1}{n}\sum_{k=1}^{n}\left(1+\frac{1}{n}\right)^k &=& \frac{1}{n}\sum_{k=1}^{n}\exp\left(k\log\left(1+\frac{1}{n}\right)\right)\\&=&\frac{1}{n}\sum_{k=1}^{n}\exp\left(\frac{k}{n}\left(1+O(1/n)\right)\right)\\&\to&\int_{0}^{1}e^{x}\,dx=\color{red}{e-1}.\end{eqnarray*}$$

Answer 2

Note that $$\sum_{k=1}^n \left(1+\dfrac1n\right)^k$$ is a geometric sum. And recall that $$\sum_{k=1}^n x^k = x \cdot \dfrac{x^n-1}{x-1}$$ Hence, we have $$\dfrac1n\sum_{k=1}^n \left(1+\dfrac1n\right)^k = \dfrac1n\left(1+\dfrac1n\right) \cdot \dfrac{(1+1/n)^n-1}{1/n} = \dfrac{n+1}n \cdot \left((1+1/n)^n-1\right)$$ Taking the limit as $n \to \infty$, gives us $e-1$.

Answer 3

If $0<x<1$ then $$ 1+k\,x<(1+x)^k<1+(k+1)\,x,\quad k\in\mathbb{N}. $$ The first inequality is obvious by the binomial theorem, while the second one is easily proved by induction false (see Winther's comment.) Then $$ \Bigl(1+\frac1n\Bigr)^k=1+\frac{x_{k,n}}{n},\quad k<x_{k,n}[<k+1\text{ this is false }] $$