# Given two series $x_n$ and $y_n$

Let $x_n = \sum_{k=n}^{\ 2n-1} \frac{1}{k}$ , $y_n = \sum_{k=n+1}^{\ 2n} \frac{1}{k}$

b) Show that $y_n \leq \ln2 \leq x_n$ for all $n$

You can approximate the sums with definite integrals from above and below (see here): $$y_n=\sum_{k=n+1}^{2n}\frac1k\le\int_n^{2n}x^{-1}\mathrm dx=\ln2+\ln n-\ln n=\ln 2$$ and $$x_n=\sum_{k=n}^{2n-1}\frac1k\ge\int_n^{2n}x^{-1}\mathrm dx=\ln 2.$$
• Okay I will try to understand. Now I want to show that both $x_n$ and $y_n$ converge to $ln2$ for $n$ going to infinity. Can I just say that, beacuse of they are limited then they converge, and the series converge if the integral converge. Is that right. May 14, 2014 at 12:12
• @AlimTeacher You can use the squeeze theorem to show that $x_n$ and $y_n$ converge to $\ln 2$ as $n\to\infty$. We have that $$\ln 2\le x_n\le \int_{n-1}^{2n-1}x^{-1}\mathrm dx=\ln\Bigl(\frac{2n-1}{n-1}\Bigr)=\ln2+\ln\Bigl(\frac{n-1/2}{n-1}\Bigr)$$ and $\ln\bigl(\frac{n-1/2}{n-1}\bigr)\to0$ as $n\to\infty$ since $\frac{n-1/2}{n-1}\to1$ as $n\to\infty$. Similarly, you can show that $y_n$ converges to $\ln 2$ as $n\to\infty$. May 14, 2014 at 12:25
• @AlimTeacher For decreasing function $f$, $$\sum_{i=a}^bf(i)\le\sum_{i=a}^b\int_{i-1}^if(s)\mathrm ds=\int_{a-1}^bf(s)\mathrm ds$$ since $f(i)\le\int_{i-1}^if(s)\mathrm ds$ (see here). May 14, 2014 at 13:08
Hint : Try to commute the left and right Riemann sum of the function $x\mapsto \frac{1}{1+x}$ on $(1,2)$.