<s>[INST] Is $x_n=\sum_{k=1}^{n^2}\frac{1}{\sqrt{n^4+k}}$ convergent? 
Let $$x_n=\sum_{k=1}^{n^2}\frac{1}{\sqrt{n^4+k}}$$
  Is $\{x_n\}$ convergent? If it is convergent then find its limit.

Here n- th term of the series $\frac{1}{\sqrt{n^4+n^2}} \to 0$ as $n \to \infty$. Can I conclude from here that $\{x_n\}$ is convergent?
Again $$\lim_{n\to \infty }x_n=\lim_{n\to \infty }\sum_{k=1}^{n^2}\frac{1}{\sqrt{n^4+k}}=\lim_{m\to \infty }\frac{1}{m} \sum_{k=1}^{m}\frac{1}{\sqrt{1+\frac{k}{m^2}}}$$Then I am stuck .Please help.
 [/INST] For each term of the sum, we have
$$
\frac{1}{\sqrt{n^4+n^2}} \leq \frac{1}{\sqrt{n^4+k}}\leq \frac{1}{\sqrt{n^4+1}}.
$$
Summing these inequalities from $k=1$ to $k=n^2$ yields
$$
\frac{1}{\sqrt{1+1/n^2}}=\frac{n^2}{\sqrt{n^4+n^2}} \leq x_n \leq \frac{n^2}{\sqrt{n^4+1}}=\frac{1}{\sqrt{1+1/n^4}} .
$$
Now both ends converge to $1$.
Therefore, by the Squeeze Theorem, $\lim_{n\rightarrow +\infty} x_n=1$.
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