Show that for $|a|<1$ the series $\sum_{n=1}^{\infty}a^nf_n$ converges in $L_2([-1,1])$.

Let $$f_n(x)=x^{n+1/2}$$ for $$n\geq 1$$ and $$x\in [-1,1]$$. Show that for $$|a|<1$$ the series $$\sum_{n=1}^{\infty}a^nf_n$$ converges in $$L_2([-1,1])$$. First, I have shown that $$\sum_{n=1}^{\infty}a^nf_n(x)$$ converges to $$\frac{ax^{3/2}}{1-ax}$$ for $$x\in [-1,1]$$. After that, I'm having difficulties to show that $$\int_{-1}^{1}\left | \frac{ax^{3/2}}{1-ax} \right |^2\,\mathrm{d}x=|a|^2\int_{-1}^{1}\frac{|x^{3}|}{\left | 1-ax \right |^2}\,\mathrm{d}x$$ is less than $$+\infty$$. The denominator in the integral is troublesome.

• Isn't there a problem with $x^{n+1/2}$ for $x<0?$ – zhw. Jan 10 at 23:35
• @zhw. You are right, I've only considered it on $[0, 1]$. I'll talk to my lecturer. – UnknownW Jan 10 at 23:40
• @zhw. It is a function from $[-1,1]$ into $\mathbb{C}$, so it's all right. For $x<0$, we may put $\sqrt{x}=i\sqrt{-x}$. – UnknownW Jan 11 at 15:33
• That's unusual, so you should probably mention it in your question. – zhw. Jan 11 at 19:42

Since $$|a|<1$$ and in the integral $$|x|<1$$, we have $$(1-ax)>0$$, so
$$|a|^2\int_{-1}^1 \frac{|x^3|}{|1-ax|^2}dx = a^2\int_0^1 \frac{x^3}{(1-ax)^2}dx\,-a^2\int_{-1}^0\frac{x^3}{(1-ax)^2}dx.$$
Substituting $$u=1-ax$$ should finish the job.
Hint: The denominator is not troublesome; it can never be $$0$$ on the given interval. Another thing: If $$\sum \|g_n\|_2 <\infty,$$ then $$\sum g_n$$ converges in $$L^2.$$
• What does $||\cdot ||_2$ mean? – UnknownW Jan 15 at 22:31
• It's the $L^2$ norm – zhw. Jan 15 at 22:58