# Series interpretation of integral

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 $$\ln$$?

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

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

• 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? – Tim Feb 1 at 17:43
• 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. – Mars Plastic Feb 1 at 17:48
• 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? – Tim Feb 1 at 18:15
• OK, so how is the integral introduced (defined) in your lecture? – Mars Plastic Feb 2 at 1:20
• 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 $\int_a^b f(x) dx = \lim\limits{n \to \infty}\int_a^b \varphi_n(x) dx$. – Tim Feb 2 at 6:53

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$$\; f(x)=\frac 1x.$$

• 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? – Tim Feb 1 at 17:29