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show sequence $a_n= \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n}$ converges to log2

my attempt:

  1. sequence $a_n$ is monotonic increasing
  2. 0<$a_n$<1/2

how to find limit?

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    $\begingroup$ Oh come on, this must be the duplicate of three duplicates. $\endgroup$ – Vincenzo Oliva Aug 14 '15 at 9:39
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$$\lim_{n\rightarrow \infty}\sum_{i=1}^{n}\frac 1 {n+i}=\lim_{n\rightarrow \infty}\sum_{i=1}^{n}\frac 1 n\frac 1 {1+\dfrac in}=\int_0^1\frac 1 {1+x}dx=\ln(1+1)-\ln(1+0)=\ln(2)$$ (Riemann sum)

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  • $\begingroup$ +1, but there should be a limit as $n \to \infty$ instead of the sum going to infinity. The sum goes fom $i=1$ to $i=n$ $\endgroup$ – 6005 Aug 14 '15 at 9:07
  • $\begingroup$ @6005 Fixed, thank you! $\endgroup$ – GeorgSaliba Aug 14 '15 at 9:09
  • $\begingroup$ ya i was hoping there would be some other way. riemann sum cannot be used to compute limit of sequence$ a_n=1+1/2+1/3...+1/n−log(n+1)$ $\endgroup$ – ketan Aug 14 '15 at 9:19
  • $\begingroup$ @ketan Then why not use the harmonic number $H_n$ that was already mentioned by @MichaelGaluza? But I suppose this becomes a slightly different question than the one you asked. $\endgroup$ – GeorgSaliba Aug 14 '15 at 9:21
  • $\begingroup$ @ketan $a_n=1+1/2+1/3+...+1/n - \log(n+1)=H_n -\log(n+1)=\log(n)+\gamma + \epsilon_n-\log(n+1)=\log(n/(n+1))+\gamma +\epsilon _n$ $\endgroup$ – GeorgSaliba Aug 14 '15 at 9:23
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$$ \begin{align} \log(2) &=\sum_{k=1}^\infty\frac{(-1)^{k-1}}k\\ &=\lim_{n\to\infty}\sum_{k=1}^{2n}\frac{(-1)^{n-1}}k\\ &=\lim_{n\to\infty}\left(\vphantom{\sum_{k=1}^n}\right.\overbrace{\sum_{k=1}^{2n}\frac1k}^{\substack{\text{sum of all}\\\text{terms}}}-2\overbrace{\sum_{k=1}^n\frac1{2k}}^{\substack{\text{sum of even}\\\text{terms}}}\left.\vphantom{\sum_{k=1}^n}\right)\\[6pt] &=\lim_{n\to\infty}\left(\sum_{k=1}^{2n}\frac1k-\sum_{k=1}^n\frac1k\right)\\[6pt] &=\lim_{n\to\infty}\sum_{k=n+1}^{2n}\frac1k\\ \end{align} $$

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  • $\begingroup$ Nice overbrace. Is there a corresponding underbrace command or somethings similar? $\endgroup$ – mathreadler Aug 14 '15 at 9:23
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    $\begingroup$ Yep. It is \underbrace :-) Then you use subscripts instead of superscripts. $\endgroup$ – robjohn Aug 14 '15 at 9:23
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Here is classical proof of it. Let $H_n=\sum_{i=1}^n 1/i$. It's well-known that $$ H_n = \ln n + \gamma + \epsilon_n, $$ where $\gamma$ is Euler–Mascheroni constant, $\epsilon_n\to 0$. So, $$ a_n = H_{2n}-H_{n} = \ln 2 + \epsilon_{2n} - \epsilon_n \to \ln2. $$

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  • $\begingroup$ what is $\epsilon_n$? is it always zero? $\endgroup$ – ketan Aug 14 '15 at 9:28
  • $\begingroup$ @ketan, no. It's sequence which tends to $0$. $\endgroup$ – Michael Galuza Aug 14 '15 at 11:39
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Notice, $$a_n=\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\ldots +\frac{1}{n+n}$$ $$=\sum_{r=1}^{\infty}\frac{1}{n+r}=\sum_{r=1}^{\infty}\frac{1}{n\left(1+\frac{r}{n}\right)}$$$$=\sum_{i=1}^{\infty}\frac{\frac{1}{n}}{1+\frac{r}{n}}$$ Now, let $\frac{r}{n}=x\implies \frac{1}{n}=dx\to 0$ $$\text{upper limit of x}=\lim_{n\to \infty}\frac{n}{n}=1$$ $$\text{lower limit of x}=\lim_{n\to \infty}\frac{1}{n}=0$$ Hence, using integration we get $$a_n=\int_{0}^{1}\frac{dx}{1+x}$$ $$=\left[\ln (1+x)\right]_{0}^{1}$$ $$=\left[\ln 2-\ln 1\right]=\ln 2$$

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