Limit of this sum

Question

Evaluate $\lim_{n\to \infty}(\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\dots+\frac{1}{n+n})$.

I converted the limit of this sum to the sum of limits. Then in each term, I divided the numerator and denominator by $n$. Each limit came out to be zero. Hence I got the answer as $0$. Have I gone wrong somewhere or is my solution correct? I have mixed feelings. Please help.

Note that for each $n,$ the sum is greater than $$1/(2n) + 1/(2n) + \cdots + 1/(2n) = n\cdot 1/(2n) = 1/2.$$ So how could the limit be $0?$ — zhw., Apr 25 '19 at 18:21
@HAMIDINESOUMARE, Yes. I know about simple series and school level integral calculus. — MrAP, Apr 25 '19 at 18:24
@zhw., But if we divide the numerator and denominator of each term by $n$, the limit is coming to be $0$. — MrAP, Apr 25 '19 at 18:25
I just showed you $$\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\dots+\frac{1}{n+n}\ge 1/2$$ for each $n.$ That shows the limit cannot be $0$. — zhw., Apr 25 '19 at 19:18

score 2 · Answer 1 · answered Apr 25 '19 at 18:49

Ok: This is inspired on Riemann Sums

Let $S_n=\sum_{k=0}^{n-1}{\frac{1}{n+k+1}}$ be your sum

We first split $[0;1]$ into $n$ intervals of the form $[\frac{k}{n};\frac{k+1}{n}]$, with $k\in\{0,1,2,...,n-1\}$

Note that if $\frac{k}{n}\le x\le \frac{k+1}{n}$ then $\frac{1}{\frac{k+1}{n}+1}\le\frac{1}{1+x}\le\frac{1}{1+\frac{k}{n}}$. Now we integrate: $$\frac{1}{k+n+1}=\int_{\frac{k}n}^{\frac{k+1}{n}}\frac{dx}{\frac{k+1}{n}+1}\le\int_{\frac{k}n}^{\frac{k+1}n}\frac{dx}{1+x}\le\int_{\frac{k}n}^{\frac{k+1}n}\frac{dx}{1+\frac{k}{n}}=\frac{1}{k+n}$$ thus: $$\sum_{k=0}^{n-1}{\frac{1}{k+n+1}}\le \sum_{k=0}^{n-1}\int_{\frac{k}n}^{\frac{k+1}n}\frac{dx}{1+x}\le\sum_{k=0}^{n-1}{\frac{1}{k+n}}$$ Which means: $$S_n\le\int_{0}^{1}{\frac{dx}{x+1}}\le S_n-\frac{1}{n}+\frac{1}{n+n}$$ Now take the limit of both side

Limit of this sum

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