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

Question

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.

Answer 1

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.

Answer 2

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 $L^2.$