- Thread starter
- #1
- Apr 13, 2013
- 3,739
Hey again!!! 
I have to check if the sequence $a_{n}=\frac{1}{\sqrt{n^2+1}}+\frac{2}{\sqrt{n^2+2}}+...+\frac{n}{\sqrt{n^2+n}}$ converges.I thought that:$$\frac{n^{2}(n+1)}{2\sqrt{n^2+n}} \leq a_{n} \leq \frac{n^{2}(n+1)}{2\sqrt{n^2+1}}$$ Because of the fact that:
$$\lim_{n \to \infty}\frac{n^{2}(n+1)}{2\sqrt{n^2+n}}=\lim_{n \to \infty}\frac{n^{2}(n+1)}{2\sqrt{n^2+1}}=\infty$$ I though that the sequence diverges.
Could you tell me if it is right?

I have to check if the sequence $a_{n}=\frac{1}{\sqrt{n^2+1}}+\frac{2}{\sqrt{n^2+2}}+...+\frac{n}{\sqrt{n^2+n}}$ converges.I thought that:$$\frac{n^{2}(n+1)}{2\sqrt{n^2+n}} \leq a_{n} \leq \frac{n^{2}(n+1)}{2\sqrt{n^2+1}}$$ Because of the fact that:
$$\lim_{n \to \infty}\frac{n^{2}(n+1)}{2\sqrt{n^2+n}}=\lim_{n \to \infty}\frac{n^{2}(n+1)}{2\sqrt{n^2+1}}=\infty$$ I though that the sequence diverges.
Could you tell me if it is right?