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I need to determine whether the following statement is true or false.

$\lim_{n\to \infty}\left( \dfrac{1}{n+1}+\dfrac{1}{n+2}+...+\dfrac{1}{n+n}\right)=\lim_{n\to \infty}\dfrac{1}{n+1}+\lim_{n\to \infty}\dfrac{1}{n+2}+...+\lim_{n\to \infty}\dfrac{1}{n+n}=0+0+...+0=0 .$

Anyone can help me? Thanks.

Sweet Cat
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  • In the first sum you've got $n$ terms each of which is greater than $\frac{1}{n+n}$. – Michael Burr Mar 19 '15 at 13:01
  • The hint (it's really an answer...and also an answer...contemplate that!) provided below is really all you need. – Daniel W. Farlow Mar 19 '15 at 13:03
  • If the number of parts of the sum on the left side was constant then you could use the fact that $\lim\limits_{n\to\infty}(a_n+b_b)=\lim\limits_{n\to\infty} a_n+\lim\limits_{n\to\infty} b_n$ if ${a_n}$ and ${b_n}$ converge. But the number of elements on the left side depends on $n$ and is increasing to the infinity so you can't use this fact. – pw1822 Mar 19 '15 at 13:05

2 Answers2

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Hint:Use a Riemman Integral $$\lim_{n\to \infty}\left( \dfrac{1}{n+1}+\dfrac{1}{n+2}+...+\dfrac{1}{n+n}\right)=\lim_{n\to \infty}\frac{1}{n}\sum_{k=1}^n\frac{1}{1+\frac{k}{n}}=\int_0^1\frac{1}{1+x}dx$$

AsdrubalBeltran
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It is not true.

Observe that for each $n\in \mathbb{N}$,

$$\dfrac{1}{n+1}+\dfrac{1}{n+2}\dots \dfrac{1}{n+n}\ge\underbrace{\dfrac{1}{n+n}+\dfrac{1}{n+n}\dots \dfrac{1}{n+n}}_{\text{n terms}}=\dfrac{n}{n+n}=\dfrac{1}{2}.$$

Therefore $\lim\limits_{n\to \infty}\left( \dfrac{1}{n+1}+\dfrac{1}{n+2}\dots \dfrac{1}{n+n}\right)\ge \lim\limits_{n\to \infty}\dfrac{1}{2}=\dfrac{1}{2} \ne0.$

ASB
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