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.
 A: 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, $\lim_{n\rightarrow +\infty} x_n=1$.
A: We can use your simplification, followed by Squeezing.
Note that 
$$1\lt \sqrt{1+\frac{k}{m^2}}\lt 1+\frac{k}{2m^2}.$$
Thus if $k\le m$, then 
$$\frac{1}{\sqrt{1+\frac{k}{m^2}}} \ge \frac{1}{1+\frac{m}{2m^2}}.$$
It is simpler not to use the simplification, and note directly that $\sqrt{n^4+k}\le \sqrt{n^4+n^2}\lt n^2+\frac{1}{2}$. 
A: Having $\lim_{x\rightarrow\infty}x_n = 0$ is a necessary condition but not a sufficient one - See the Harmonic series for ex, $\sum \frac{1}{n}$ which doesn't converge.
I know it's a clumsy way to do it, but would the following be good?
$$\lim_{L\rightarrow\infty} \int_1^L \frac{1}{\sqrt{L^2+x}}
= \lim_{L\rightarrow\infty} 2\left( \sqrt{L^2+L} - \sqrt{L^2+1} \right)
= \lim_{L\rightarrow\infty} 2\frac{L-1}{\sqrt{L^2+L} + \sqrt{L^2+1}} = 1$$
and conclude that by the integral test it converges?
