Series interpretation of integral

Question

I'm currently stuck with the following question:

Prove, that $\ln(2) = \lim\limits_{n \to \infty} \sum_{k=1}^n \frac{1}{n+k}$ by rewriting the left side as an Integral.

So my current thoughts are:

$\ln(2) = \int_1^2 \frac{1}{x} \mathrm{d}x$

$\lim\limits_{n \to \infty} \sum_{k=1}^n \frac{1}{n+k} = \lim\limits_{n \to \infty} \sum_{k=1}^n \frac{1}{n} \frac{1}{\frac{k}{n} + 1} = \lim\limits_{n \to 0} \sum_{k=1}^{\frac{2-1}{n}} n \frac{1}{nk + 1}$

Like this, the sum is rewritten as a step function with the width of each equidistant step decreasing. If the sum is the upper sum, I have to show, that $\frac{1}{nk+1}=\ln(1 + \frac{k+1}{n})$. How can I "get rid" if the $\ln$?

Answer 1

You're almost there, just use the approximation of the integral by a Riemann sum over the grid $1, 1+ 1/n,\ldots, 2$, then

$$\lim\limits_{n \to \infty} \sum_{k=1}^n \frac{1}{n} \frac{1}{\frac{k}{n} + 1}= \lim\limits_{n \to \infty} \sum_{k=1}^n \left[\left(1+\frac{k}{n}\right)-\left(1+\frac{k-1}{n}\right)\right] \frac{1}{1+\frac{k}{n}}=\int_1^2\frac{1}{x}dx.$$

Answer 2

hint

Write it as

$$\frac{b-a}{n}\sum_{k=1}^nf\left(a+k\frac{b-a}{n}\right)$$

the limit will be $$\int_a^bf(x)dx$$

In your case, $a=0\; \; b=1, f(x)=\frac{1}{1+x}$ Or $a=1,\;b=2 \;$ and$ \; f(x)=\frac 1x.$