limit of the sum $\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n} $ [duplicate]

Solution 1:

$$\int_{k}^{k+1} \frac{1}{x}dx \leq \dfrac{1}{k} \leq \int_{k-1}^{k} \frac{1}{x}dx.$$ $$ \ln\frac{2n+1}{n} \leq \sum_{k=n}^{2n}\frac{1}{k} \leq \ln\frac{2n}{n-1}. $$

Solution 2:

We are going to use the Euler's constant $$\lim_{n\to\infty}\left(\left(1+\frac{1}{2}+\cdots+\frac{1}{2n}-\ln (2n)\right)-\left(1+\frac{1}{2}+\cdots+\frac{1}{n}-\ln n\right)\right)=\lim_{n\to\infty}(\gamma_{2n}-\gamma_{n})=0$$

Hence the limit is $\ln 2$.

Solution 3:

I thought it might be instructive to present an approach that uses straightforward arithmetic and application of Leibniz's Test for alternating series. To that end, we now proceed.

It is not difficult to show that

$$\sum_{k=n+1}^{2n} \frac{1}{k}=\sum_{k=1}^{2n }\frac{(-1)^{k-1}}{k}\tag1$$

To show the validity of $(1)$, we simply write

$$\begin{align} \sum_{k=n+1}^{2n}\frac1k&=\sum_{k=1}^{2n}\frac1k-\sum_{k=1}^n \frac1k\\\\ &=\sum_{k=1}^n\left(\frac1{2k}+\frac1{2k+1}\right)-2\sum_{k=1}^n\frac1{2k}\\\\ &=\sum_{k=1}^n\left(\frac1{2k+1}-\frac1{2k}\right)\\\\ &=\sum_{k=1}^{2n}\frac{(-1)^{k-1}}{k}\tag2 \end{align}$$

Notice that the sum on the right-hand side is the alternating harmonic series, which converges as guaranteed by Leibniz's test. And we conclude that the series converges!

BONUS: EVALUATING THE LIMIT

In fact, using the Taylor series of $\log(1+x)=\sum_{k=1}^\infty \frac{(-1)^{k-1}x^k}{k}$, and evaluating it at $x=1$, we find that the series of interest is equal to $\log(2)$. Hence, we find that

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

And we are done.