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

Question

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?

you can't take the product of the limits as you did in step 3 — Alex, Jun 16 '14 at 11:57
Squeezing is very good. What is the smallest term in the sum? — Daniel Fischer, Jun 16 '14 at 12:00
So which bounds do you get for the sum if you use the largest and smallest term to estimate? — Daniel Fischer, Jun 16 '14 at 12:02
the limit $\frac{1}{\sqrt{1+\frac{i}{n^2}}}=1$, so the sum diverges — Alex, Jun 16 '14 at 12:03

score 10 · Accepted Answer · answered Jun 16 '14 at 12:15

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.

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

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