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Prove limit of $\displaystyle \frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{n+n}$ exists and lies between $0$ and $1$.

So far I have $\displaystyle \lim_{n\rightarrow\infty}\sum_{k=1}^n\frac{1}{k+n}=L$ for some $L>0$.

Then given any $\epsilon >0,\exists N>0$ such that if $n>N$, then $\displaystyle\left|\sum_{k=1}^n\frac{1}{k+n}-L\right|<\epsilon$

A hint would be appreciated!

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marked as duplicate by Martin Sleziak, N. F. Taussig, John Gowers, David K, apnorton Mar 30 '15 at 14:41

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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It's a monotone rising sequence: if $$s_n = \frac{1}{n+1} + ... + \frac{1}{2n},$$ then we have $$s_{n+1} - s_n = \frac{1}{2n+2} + \frac{1}{2n+1} - \frac{1}{n+1} = \frac{1}{2n+1} - \frac{1}{2n+2} > 0.$$

It is bounded below by $0$; and it's bounded above by $1$, because $$s_n = \frac{1}{n+1} + ... + \frac{1}{2n} < \frac{1}{n} + ... + \frac{1}{n} = 1.$$ So it converges to a limit between $0$ and $1$.

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  • $\begingroup$ I like the use of the monotone convergence theorem here. $\endgroup$ – lightfish Sep 19 '13 at 0:56
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Hint: It's a Riemann sum. ${}$${}$

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  • $\begingroup$ Thank you, I did try Riemann sums, and got $$\displaystyle\lim\frac{1}{n}\sum_{k=1}^n\frac{1}{\frac{k}{n}+1}$$. I know Riemann sum to integral is of the form $$\sum_{i=0}^{n-1}f(t_i)\delta x_i=\int_a^bf(x)dx$$, where $\delta x_i=\frac{b-a}{n}$. In this case, would $f(t)=\frac{1}{\frac{x}{n}+1}$? $\endgroup$ – lightfish Sep 19 '13 at 0:19
  • $\begingroup$ ..."and got it right" ...or what? $\endgroup$ – DonAntonio Sep 19 '13 at 0:20
  • $\begingroup$ @DonAntonio The suspense, it's killing me :) $\endgroup$ – PeterM Sep 19 '13 at 0:21
  • $\begingroup$ sorry, whenever I hit "enter" it saved my comment... $\endgroup$ – lightfish Sep 19 '13 at 0:22
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    $\begingroup$ @brenna $\large{\tt shift}$ $+$ $\large{\tt enter}$ goes to a new line without publishing your comment. $\endgroup$ – Felix Marin Sep 19 '13 at 0:27
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Note that $$ \begin{align} \sum_{k=n+1}^{2n}\frac1k &=\sum_{k=1}^{2n}\frac1k\hphantom{\frac1{2k-1}+}-\sum_{k=1}^n\frac1k\\ &=\sum_{k=1}^n\frac1{2k-1}+\frac1{2k}-\frac1k\\ &=\sum_{k=1}^n\frac1{2k-1}-\frac1{2k}\\ &=\sum_{k=1}^{2n}(-1)^{k-1}\frac1k\tag{1} \end{align} $$ and by the Alternating Series Test, the series in $(1)$ converges.

Also by the Alternating Series Test, the limit is between any two consecutive partial sums; in particular, the first two are $0$ and $1$ (followed by $\frac12$, $\frac56$, and $\frac7{12}$).

In this answer, that limit is shown, without calculus, to be $\log(2)$.


Evaluating the Limit

We will use the inequality $$ e^x\ge1+x\tag{2} $$ Substituting $x\mapsto-x$ in $(2)$ and taking reciprocals yields $$ e^x\le\frac1{1-x}\tag{3} $$ Applying $(2)$ gives $$ \begin{align} e^{\frac1{n+1}+\frac1{n+2}+\cdots+\frac1{2n}} &\ge\frac{n+2}{n+1}\frac{n+3}{n+2}\cdots\frac{2n+1}{2n}\\[6pt] &=\frac{2n+1}{n+1}\\[4 pt] &=2-\frac1{n+1}\tag{4} \end{align} $$ Applying $(3)$ gives $$ \begin{align} e^{\frac1{n+1}+\frac1{n+2}+\cdots+\frac1{2n}} &\le\frac{n+1}n\frac{n+2}{n+1}\cdots\frac{2n}{2n-1}\\[6pt] &=\frac{2n}n\\[9 pt] &=2\tag{5} \end{align} $$ Therefore, $$ \log\left(2-\frac1{n+1}\right)\le\frac1{n+1}+\frac1{n+2}+\cdots+\frac1{2n}\le\log(2)\tag{6} $$ Thus, the limit is $\log(2)$.

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  • $\begingroup$ Is the $H(2n)-H(n)$ a trick for series used in conjunction with alternating series test? I'm going to have to study the linked answer a bit. $\endgroup$ – lightfish Sep 19 '13 at 0:52
  • $\begingroup$ @brenna: I think I've brought everything into this answer, so you shouldn't need to read the previous one now. $\endgroup$ – robjohn Sep 19 '13 at 0:54
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Let $\xi_{k} \equiv k/n$ and $\Delta\xi \equiv \xi_{k + 1} - \xi_{k} = 1/n$

$$ \sum_{k = 1}^{n}{1 \over k + n} = \sum_{k = 1}^{n}{1 \over n\xi_{k} + n}\,n\Delta\xi \sim \int_{1/n}^{1}{{\rm d}\xi \over \xi + 1} = \ln\left(2\right) - \ln\left(1 + {1 \over n}\right) \to \color{#ff0000}{\large\ln\left(2\right)} $$

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$f(n)=\sum\limits_{k=1}^n \frac{1}{n+k}$

$f(n+1)-f(n)>0$

$\lim\limits _{n\to \infty} f(n)=\log2<1$

$f(2)=\frac{7}{12}>0$

Hence $\forall n \in \mathbb{N}^+,0<f(n)<1$

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The sum is increasing and bounded.$$\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{n+n}<\frac{1}{n}+\frac{1}{n}+\ldots+\frac{1}{n} =1.$$

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