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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 $\ln$?

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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

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

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  • $\begingroup$ So $\left[\left(1+\frac{k}{n}\right)-\left(1+\frac{k-1}{n}\right)\right]$ is the "width" of the subdivision and $\frac{1}{1+ \frac{k}{n}}$ is the "height" respectively. But where do I know from, that this infinite sum is equal to the Integral? $\endgroup$ – Tim Feb 1 at 17:43
  • $\begingroup$ That's the definition of the Riemann integral. And be careful: It's not an infinite sum, but the limit of a sequence of finite sums. $\endgroup$ – Mars Plastic Feb 1 at 17:48
  • $\begingroup$ Ok, thank you very much. We haven't defined the Riemann sum in our lecture yet, is there a way of solving the problem without this definition? $\endgroup$ – Tim Feb 1 at 18:15
  • $\begingroup$ OK, so how is the integral introduced (defined) in your lecture? $\endgroup$ – Mars Plastic Feb 2 at 1:20
  • $\begingroup$ We have introduced the integral with step functions: Let $\varphi$ be a step function, then the integral of $\varphi$ over $[a,b]$ is defined as $\int_a^b \varphi(x) dx = \sum_{i=1}^n c_i (x_i - x_{i-1})$. And we have the definition of the integral of a regulated function: If $(\varphi_n)$ is uniformly converging to $f$, then $\int_a^b f(x) dx = \lim\limits{n \to \infty}\int_a^b \varphi_n(x) dx$. $\endgroup$ – Tim Feb 2 at 6:53
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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.$

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  • $\begingroup$ So it would be $\int_1^2 \frac{1}{x} dx = \frac{1}{n} \sum_{k=1}^n \frac{1}{1 + \frac{k}{n}}$ which is equal to my upper transformation. But how can I prove the equivalence you provided to me? $\endgroup$ – Tim Feb 1 at 17:29

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