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

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 $\log(2)$.

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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 $\log(2)$.