3
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Direct way leads to nowhere.

$$\lim_{n\rightarrow \infty} \sum^{n}_{i=1} \frac{1}{\sqrt {n^2 +i} } = \lim_{n\rightarrow \infty} \sum^{n}_{i=1} \frac{1}{n \sqrt {1 +\frac{i}{n^2}} } = $$

$$ = \lim_{n\rightarrow \infty} \frac{1}{n} \cdot \lim_{n\rightarrow \infty} \sum^{n}_{i=1} \frac{1}{\sqrt {1 +\frac{i}{n^2}} } = 0 \cdot \sum^{n}_{i=1} 1 = 0 \cdot n $$

So, I tried to apply squeeze theorem:

$$\lim_{n\rightarrow \infty} \sum^{n}_{i=1} \frac{1}{\sqrt {n^2 +i} } \leq \lim_{n\rightarrow \infty} \sum^{n}_{i=1} \frac{1}{n} = 1$$

Could you help me to figure out the lower bound?

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  • $\begingroup$ you can't take the product of the limits as you did in step 3 $\endgroup$ – Alex Jun 16 '14 at 11:57
  • $\begingroup$ Why? Both limits are defined. $\endgroup$ – user Jun 16 '14 at 11:58
  • $\begingroup$ are you sure the denominator is not $\sqrt{n^2-i^2}$? $\endgroup$ – Alex Jun 16 '14 at 11:58
  • 1
    $\begingroup$ Squeezing is very good. What is the smallest term in the sum? $\endgroup$ – Daniel Fischer Jun 16 '14 at 12:00
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    $\begingroup$ the limit $\frac{1}{\sqrt{1+\frac{i}{n^2}}}=1$, so the sum diverges $\endgroup$ – Alex Jun 16 '14 at 12:03
9
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Notice that $$\frac{n}{\sqrt{n^2+n}} =\sum_{i=1}^n \frac{1}{\sqrt{n^2+n}}\leq \sum_{i=1}^n\frac{1}{\sqrt{n^2+i}}\leq \sum_{i=1}^n \frac{1}{n} =1$$ Since $\frac{n}{\sqrt{n^2+n}}=\frac{1}{\sqrt{1+\frac{1}{n}}}\to 1$, you get that $1$ as your limit by squeeze theorem.

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