Calculating the limit of $\lim\limits_{n \to \infty} (\frac{1}{n+1} + \frac{1}{n+2} + \frac{1}{n+3} + ...+\frac{1}{2n})$

Question

Hello everyone how can I calculate the limit of:

$\lim\limits_{n \to \infty} (\frac{1}{n+1} + \frac{1}{n+2} + \frac{1}{n+3} + ...+\frac{1}{2n})$

Answer 1

It is a very classical question on mathstack... please do some searches before asking a question !

As an answer, consider this :

$$ \sum_{n=k+1}^{2k}\frac1n= \frac 1 n \sum_{n=1}^{k} \frac1 {{ 1+ \frac n k} } $$

which is related to this integral via a Riemann sum :

$$\int_1^2\frac1x\,dx=\ln 2.$$

Answer 2

Hint:

Factor out $n$ in all denominators: $$\frac{1}{n+1} + \frac{1}{n+2} + …+\frac{1}{2n}=\frac1n\biggl(\frac1{1+\frac1n}+\frac1{1+\frac2n}\dots+\frac1{1+\frac nn}\biggr)$$ and recognise a Riemann sum.

Answer 3

I thought it might be instructive to present an approach that avoids appealing to integrals. Rather, we make use of elementary arithmetic and the Taylor series for $\log(1+x)=\sum_{k=1}^\infty\frac{(-1)^{k-1}x^k}{k}$. To that end, we proceed.

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Note that we can write the sum of interest as

$$\begin{align} \sum_{k=1}^n\frac{1}{n+k}&=\sum_{k=n+1}^{2n}\frac{1}{k}\\\\ &=\sum_{k=1}^{2n}\frac1k-\sum_{k=1}^n\frac1k\\\\ &=\sum_{k=1}^{n}\left(\frac{1}{2k-1}+\frac1{2k}\right)-\sum_{k=1}^n\frac1k\\\\ &=\sum_{k=1}^{2n}\frac{(-1)^{k-1}}{k} \end{align}$$

Then, using the Taylor series for $\log(1+x)$, evaluated at $x=1$, we find

$$\lim_{n\to\infty}\sum_{k=1}^\infty \frac1{n+k}=\log(2)$$

And we are done!