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 $n$
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asked May 14, 2014 at 11:09
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2 Answers
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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 $$ x_n=\sum_{k=n}^{2n-1}\frac1k\ge\int_n^{2n}x^{-1}\mathrm dx=\ln 2. $$
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$\begingroup$ 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. $\endgroup$ Commented May 14, 2014 at 12:12
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$\begingroup$ @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 $n\to\infty$. $\endgroup$– Cm7F7BbCommented May 14, 2014 at 12:25
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$\begingroup$ okay, but why do you choose the lower limit to be n-1 $\endgroup$ Commented May 14, 2014 at 12:51
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Hint : Try to commute the left and right Riemann sum of the function $x\mapsto \frac{1}{1+x}$ on $(1,2)$.
answered May 14, 2014 at 11:12