Prove that $f(x) = \sum_{n=1}^{\infty} {\frac{x^2 \sqrt{n} + 1 }{ 2n^2 + x } } $ is continuous in $(-1,2)$. I was asked this question:

Prove that $f(x) = \sum_{n=1}^{\infty} {\frac{x^2 \sqrt{n} + 1  }{ 2n^2 + x   } } $ is continuous in $(-1,2)$, when $x \in \mathbb R$.

On first thought, I said that if we prove that the summation uniformly converges, and it is obvious that the series of functions $f_n(x)$ are continuous, then according to uniform convergence rules, $f(x)$ is continuous everywhere, especially in $(-1,2)$.
Now I wanted to prove that it is uniformly convergent. By trying to apply the Weistrass M-Test I only have been able to say that it is $\leq |\frac{x^2}{x}| = |x|$ which helped with nothing. Then I thought about finding the general $g(x)$ that it might converge to and check if $a_n = \lim_{n \to \infty} \sup |g_n(x) - g(x)| = 0$ but I couldn't find what $g_n(x)$ was since I got to a point where I get $\frac{0}{\infty}$.
Any help will be appreciated, thanks!
 A: You are correct in your approach. The series does converge uniformly on the interval $(-1,2)$. As you implied correctly, since each term of the given series defines a function that is continuous on the interval $(-1,2)$, it follow that the series defines a continuous function on $(-1,2)$.
And, you can use the $M$-test to prove the convergence is indeed uniform on $(-1,2)$.  However, your bound is not suitable.  
But, if you notice that the terms of the series are positive, then an upper bound for $f_n(x)={x^2\sqrt n+1\over 2n^2+x}$, valid for all $x\in(-1,2)$, can be obtained simply by replacing the numerator by its largest value in the interval $(-1,2)$ and the denominator by its smallest value in $(-1,2)$. This  leads to the bound $M_n={4\sqrt n+1\over 2n^2-1}$ for $f_n$. To emphasise, for each positive integer $n$, we have $|f_n(x)|\le M_n$ for all $x\in(-1,2)$.
Fortunately, this rather sloppy approximation works for the $M$-test:
The only thing left to do is to show that $\sum\limits_{n=1}^\infty M_n$ converges. This can be done by applying the Limit Comparision Test to this series and the convergent $p$-series  $\sum\limits_{n=1}^\infty{1\over n^{3/2}}$.
