# Prove limit of $\frac1{n+1}+\frac1{n+2}+\ldots+\frac1{n+n}$ exists and lies between $0$ and $1$. [duplicate]

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!

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

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 $1$.
Hint: It's a Riemann sum. {}$${} • Thank you, I did try Riemann sums, and got$$\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$$, where \delta x_i=\frac{b-a}{n}. In this case, would f(t)=\frac{1}{\frac{x}{n}+1}? – lightfish Sep 19 '13 at 0:19 • ..."and got it right" ...or what? – DonAntonio Sep 19 '13 at 0:20 • @DonAntonio The suspense, it's killing me :) – PeterM Sep 19 '13 at 0:21 • sorry, whenever I hit "enter" it saved my comment... – lightfish Sep 19 '13 at 0:22 • @brenna \large{\tt shift} + \large{\tt enter} goes to a new line without publishing your comment. – Felix Marin Sep 19 '13 at 0:27 Note that$$ \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} $$Thus, the limit is \log(2). • Is the H(2n)-H(n) a trick for series used in conjunction with alternating series test? I'm going to have to study the linked answer a bit. – lightfish Sep 19 '13 at 0:52 • @brenna: I think I've brought everything into this answer, so you shouldn't need to read the previous one now. – robjohn Sep 19 '13 at 0:54 Let \xi_{k} \equiv k/n and \Delta\xi \equiv \xi_{k + 1} - \xi_{k} = 1/n$$ \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)} $$f(n)=\sum\limits_{k=1}^n \frac{1}{n+k} f(n+1)-f(n)>0 \lim\limits _{n\to \infty} f(n)=\log2<1 f(2)=\frac{7}{12}>0 Hence \forall n \in \mathbb{N}^+,0<f(n)<1 The sum is increasing and bounded.$$\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{n+n}<\frac{1}{n}+\frac{1}{n}+\ldots+\frac{1}{n} =1.\$