Show that $f(x):=\sum_{n=1}^{\infty}\frac{x}{(1+nx^{2})n^{\alpha}}\rightarrow 0$ as $x\rightarrow 0$, if $\alpha>\frac{1}{2}$. Consider the function defined by $$f(x):=\sum_{n=1}^{\infty}\frac{x}{(1+nx^{2})n^{\alpha}}.$$
I have showed that, by Weierstrass M-test, when $\alpha>\frac{1}{2}$, the series converges uniformly to $f$ on $[\epsilon, R]$ for $\epsilon, R>0$ arbitrarily fixed. Hence, $f(x)$ is continuous on $(0,\infty)$ for $\alpha>\frac{1}{2}$.
The second part of this exercise asks me to prove that

For $\alpha>\frac{1}{2}$, $$f(x)\longrightarrow 0\ \text{as}\ x\rightarrow 0.$$

Since I only showed that $f(x)$ is continuous on $(0,\infty)$, not at $x=0$, I cannot use $\lim_{x\rightarrow 0}f(x)=f(0).$
What alternative do I have? I tried to show that the series converges uniformly on $[-\epsilon, \epsilon]$ so that taking $\epsilon\searrow 0$, we can conclude that $f(x)$ is continuous at $0$. However I failed since the bound in M-test cannot be established..
Is there any other way to approach this exercise? Thank you!
Edit 1:
Okay I figured it out. It turned out that the proof is really complicated.  Note that there is no way to prove the uniform convergence in either $[0,\epsilon]$ or $[-\epsilon,\epsilon]$. The second one cannot give you the desired $M_{n}$ in Weierstrass $M-$test. The first one give you $M_{n}=\frac{\epsilon}{2^{\alpha}}$ but $\alpha>\frac{1}{2}$ so the series diverges.
Give me several minutes so that I can post the solution in the answer my post section.
 A: Note for $x\ge 0$, we have from the AM-GM inequality
$$\left|\frac{x}{(1+nx^2)n^\alpha}\right|\le \frac{1}{2n^{\alpha+1/2}}$$
Inasmuch as for $\alpha>1/2$, $\sum_{n=1}\frac1{n^{\alpha+1/2}}<\infty$, we can apply the Dominated Convergence Theorem (note that we also have uniform convergence for $x\in [0,\infty)$) to find
$$\lim_{x\to 0}\sum_{n=1}^\infty \frac{x}{(1+nx^2)n^\alpha}=\sum_{n=1}^\infty \lim_{x\to 0}\frac{x}{(1+nx^2)n^\alpha}0$$
Now, repeat for the case $x\le 0$.
A: Define $g(x,t):=\frac{x}{(1+tx^{2})t^{\alpha}}$ for $x\in (0,\infty)$ fixed and $t=1,2,3,\cdots$. Note that $g(x,t)$ is decreasing in $t$, and thus for $t\in [n,n+1]$, $n=1,2,3,\cdots$, using the monotonicity of integral, we have $$g(x,n+1)=g(x,n+1)\int_{n}^{n+1}1dt=\int_{n}^{n+1}g(x,n+1)dt\leq \int_{n}^{n+1}g(x,t)dt\leq\int_{n}^{n+1}g(x,n)dt=g(x,n).$$
Therefore, we have $$\dfrac{x}{(1+(n+1)x^{2})(n+1)^{\alpha}}\leq \int_{n}^{n+1}g(x,t)dt\leq\dfrac{x}{(1+nx^{2})n^{\alpha}}.$$ Summing oer $n=1,2,\cdots,$ we have $$f(x)-\dfrac{x}{1+x^{2}}\leq \int_{1}^{\infty}\dfrac{x}{(1+tx^{2})t^{\alpha}}dt\leq f(x).$$
Taking $x\rightarrow 0$, we see that $\frac{x}{1+x^{2}}\longrightarrow 0$, hence we have $$\lim_{x\rightarrow 0}\int_{1}^{\infty}\dfrac{x}{(1+tx^{2})t^{\alpha}}dt\leq\lim_{x\rightarrow 0}f(x)\leq \lim_{x\rightarrow 0}\int_{1}^{\infty}\dfrac{x}{(1+tx^{2})t^{\alpha}}dt.$$
Hence, $$\lim_{x\rightarrow 0}f(x)= \lim_{x\rightarrow 0}\int_{1}^{\infty}\dfrac{x}{(1+tx^{2})t^{\alpha}}dt.$$
To evaluate the RHS, replace $u:=\sqrt{t}$ (this may take a while to find), so that $$\lim_{x\rightarrow 0}\int_{1}^{\infty}\dfrac{x}{(1+tx^{2})t^{\alpha}}dt=\lim_{x\rightarrow 0}2\int_{1}^{\infty}\dfrac{x}{(1+u^{2}x^{2})u^{2\alpha-1}}du.$$ But not that $$\Big|\dfrac{x}{(1+u^{2}x^{2})u^{2\alpha-1}}\Big|\leq x^{-1}u^{-2\alpha-1}$$ which satisfies $\int_{1}^{\infty}x^{-1}u^{-2\alpha-1}du<\infty.$
Therefore, it follows from dominated convergence theorem that $$\lim_{x\rightarrow 0}2\int_{1}^{\infty}\dfrac{x}{(1+u^{2}x^{2})u^{2\alpha-1}}du= 2\int_{1}^{\infty}\lim_{x\rightarrow 0}\dfrac{x}{(1+u^{2}x^{2})u^{2\alpha-1}}du=0.$$
Thus, $\lim_{x\rightarrow 0}f(x)=0.$

Please let me know if the proof has typos or mistakes. I omit several computations since it takes time to write down.
