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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.2011-11-20
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    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."2011-11-20
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    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?2011-11-20
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    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.2011-11-20
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    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?2011-11-20
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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