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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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I don't really feel like it right now. – Tyler Nov 20 '11 at 1:09
The point Tyler is trying to make is that you shouldn't state your questions as commands. It's better to ask "How can I prove this" instead of saying "Prove this." – Aaron Nov 20 '11 at 1:10
Hint: There are $n$ terms, so $1>n/\sqrt{n^2+1}>v_n>n/\sqrt{n^2+n}$. What are the limits of the two sides of the inequality? – Aaron Nov 20 '11 at 1:13
Which grammar you use is really not relevant. What is relevant is that you're just copying an exercise without adding any thoughts of your own. That will make many members much less likely to want to help you, even if you rephrase the question with different grammar but no new content. – Henning Makholm Nov 20 '11 at 1:13
The story does not end at the command Aaron mentioned: some people even consider good manners to explain what you know, what you tried and why it failed. Amazing, no? – Did Nov 20 '11 at 1:14
up vote 1 down vote accepted

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

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

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

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! – FrConnection Nov 20 '11 at 12:08

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