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!
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. ${}$${}$
Note that $$ \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} $$ Thus, the limit is $\log(2)$.
Let $\xi_{k} \equiv k/n$ and $\Delta\xi \equiv \xi_{k + 1} - \xi_{k} = 1/n$
$$ \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)} $$
The sum is increasing and bounded.$$\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{n+n}<\frac{1}{n}+\frac{1}{n}+\ldots+\frac{1}{n} =1.$$
$f(n)=\sum\limits_{k=1}^n \frac{1}{n+k}$
$f(n+1)-f(n)>0$
$\lim\limits _{n\to \infty} f(n)=\log2<1$
$f(2)=\frac{7}{12}>0$
Hence $\forall n \in \mathbb{N}^+,0<f(n)<1$
