Does $a_{n} = \frac{1}{\sqrt{n^2+n}} + \frac{1}{\sqrt{n^2+n+1}} + ... + \frac{1}{\sqrt{n^2+2n-1}}$ converge? $a_{n} = \frac{1}{\sqrt{n^2+n}} + \frac{1}{\sqrt{n^2+n+1}} + ... +  \frac{1}{\sqrt{n^2+2n-1}}$
and I need to check whether this sequence converges to a limit without finding the limit itself. I think about using the squeeze theorem that converges to something (I suspect '$1$').
But I wrote $a_{n+1}$ and $a_{n-1}$ and it doesn't get me anywhere...
 A: $$\frac{1}{\sqrt{n^2+2n-1}}+\frac{1}{\sqrt{n^2+2n-1}}\cdots+\frac{1}{\sqrt{n^2+2n-1}}\le\frac{1}{\sqrt{n^2+n}} + \frac{1}{\sqrt{n^2+n+1}} + \cdots +  \frac{1}{\sqrt{n^2+2n-1}}\le\frac{1}{\sqrt{n^2+n}}+\frac{1}{\sqrt{n^2+n}}\cdots+\frac{1}{\sqrt{n^2+n}}$$
$$\frac{n}{\sqrt{n^2+2n-1}}\le\frac{1}{\sqrt{n^2+n}} + \frac{1}{\sqrt{n^2+n+1}} + \cdots +  \frac{1}{\sqrt{n^2+2n-1}}\le\frac{n}{\sqrt{n^2+n}}$$
$$\lim_{n\rightarrow\infty}\frac{n}{\sqrt{n^2+2n-1}}\le\lim_{n\rightarrow\infty}\frac{1}{\sqrt{n^2+n}} + \frac{1}{\sqrt{n^2+n+1}} + \cdots +  \frac{1}{\sqrt{n^2+2n-1}}\le\lim_{n\rightarrow\infty}\frac{n}{\sqrt{n^2+n}}$$
$$\lim_{n\rightarrow\infty}\frac{n}{n\sqrt{1+\frac2n-\frac1{n^2}}}\le\lim_{n\rightarrow\infty}\frac{1}{\sqrt{n^2+n}} + \frac{1}{\sqrt{n^2+n+1}} + \cdots +  \frac{1}{\sqrt{n^2+2n-1}}\le\lim_{n\rightarrow\infty}\frac{n}{n\sqrt{1+\frac1n}}$$
$$1\le\lim_{n\rightarrow\infty}\frac{1}{\sqrt{n^2+n}} + \frac{1}{\sqrt{n^2+n+1}} + \cdots +  \frac{1}{\sqrt{n^2+2n-1}}\le1$$
$$\lim_{n\rightarrow\infty}\frac{1}{\sqrt{n^2+n}} + \frac{1}{\sqrt{n^2+n+1}} + \cdots +  \frac{1}{\sqrt{n^2+2n-1}}=1$$
A: You're right, you just have to consider that
$$\frac{n}{\sqrt{n^2+n}}\ge a_{n} \ge  \frac{n}{\sqrt{n^2+2n-1}}$$
Then take the limit and you're done. 
A: On the one hand, 
$$a_n \ge \mbox{smallest summand} \times \mbox{number of summands}= \frac{1}{\sqrt{n^2+2n-1}}\times  n .$$
To deal with the denominator, observe that  
$$n^2+2n-1 \le n^2+2n+1=(n+1)^2.$$
On the other hand, 
$$a_n \le \mbox{largest  summand}\times \mbox{number of summands} = \frac{1}{\sqrt{n^2+n}}\times n.$$ 
To deal with the denominator, observe that 
$$n^2+n \ge n^2$$. 
