<s>[INST] How can I check the convergence of the sequence? Does it diverge? How can I check the convergence of the sequence $\frac{1}{\sqrt{n^2+1}}+\frac{2}{\sqrt{n^2+2}}+\cdots+\frac{n}{\sqrt{n^2+n}}$? I think that it diverges,because it is bounded below from $\frac{n(n+1)}{2\sqrt{n^2+n}} $ and above from $\frac{n(n+1)}{2\sqrt{n^2+1}}$..Is this correct?
 [/INST] Since $n^2\le n^2+k\le\left(n+\frac12\right)^2$ for $0\le k\le n$, we have
$$
\frac1{n+\frac12}\sum_{k=1}^nk\le\sum_{k=1}^n\frac{k}{\sqrt{n^2+k}}\le\frac1n\sum_{k=1}^nk
$$
Thus,
$$
\frac n2\le\frac{n(n+1)}{2n+1}\le\sum_{k=1}^n\frac{k}{\sqrt{n^2+k}}\le\frac{n+1}2{}
$$
That is, the sequence diverges.
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