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Good day, I would like to solve this problem using the theorem of the comparison, but I do not understand how to proceed, can anyone help me?

How do I prove that $(v_n)$, defined by $v_n:= \frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots +\frac{1}{\sqrt{n^2+n}},$ is convergent?

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    @djmaxwell Do you know of the squeeze theorem?2011-11-20

1 Answers 1

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$\frac{n}{\sqrt{n^2+n}} = \frac{1}{\sqrt{n^2+n}}+\frac{1}{\sqrt{n^2+n}}+\cdots+\frac{1}{\sqrt{n^2+n}} \leq $

$\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots +\frac{1}{\sqrt{n^2+n}}$

$\leq \frac{1}{\sqrt{n^2}}+\frac{1}{\sqrt{n^2}}+\cdots+\frac{1}{\sqrt{n^2}} = \frac{n}{\sqrt{n^2}} = 1$

The inequalities hold since bigger denominators make smaller fractions and obviously: $n^2+n \geq n^2+i \geq n^2$ for $i=1,\dots,n$.

Next, $\lim\limits_{n\to\infty} \frac{n}{\sqrt{n^2+n}}= 1$. So your summation is squeezed between 1 and 1. Thus its limit is 1.

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    very good! tanks clear!2011-11-20