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Problem: Calculate limit of $\frac{1}{\sqrt{n+1}} + \frac{1}{\sqrt{n+2}} + \frac{1}{\sqrt{n+3}} + \cdots + \frac{1}{\sqrt{n+n^2}}$ as $n$ approaches infinity.

Solution: Denote the above some as $X$, then we can bound it: $$ \infty\longleftarrow\frac{1}{\sqrt{n+n^2}} \lt X \lt \frac{n^2}{\sqrt{n+1}} \lt \frac{n^2}{\sqrt{n}} = \sqrt{\frac{n^4}{n}}\longrightarrow \infty.$$

So, from the Squeeze Principle, $\lim X = \infty$. Am I doing the right thing?

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    I had assumed the first fraction in the inequality was a typo when I TeX-ed up the question; since several responders commented, I've restored it (after TeX-ing) to its original statement.2010-11-17

5 Answers 5