Prove that the function $L^2$ converges Prove that the series
$$\sum_{n=0}^\infty (-1)^n x^{2n}$$
$L^2$ converges in $(-1,1)$.
I thought that we needed a target function since the definition of $L^2$ convergence is that 
$$\lim_{N\to \infty}\int_a^b |f(x) - \sum_{n=0}^Nf(x_n)|^2 dx = 0$$
but this problem has no target function
 A: $\sum_{n=0}^{N}(-1)^{n} x^{2n}=\frac {1-(-x^{2})^{N+1}} {1+x^{2}}$. Hence the partial sums of the series tend to $\frac 1 {1+x^{2}}$ and they are dominated by $\frac 2 {1+x^{2}}$.  By Dominated Convergence Theorem the series converges in $L^{2}(-1,1)$. 
A: Hint. There is a target function $f\in L^2(-1,1)$. For $x\in(-1,1)$ we have a convergent geometric series:
$$f(x)=\sum_{n=0}^\infty (-1)^n x^{2n}=\frac{1}{1+x^2}.$$
A: HINT: you dont need a target function. Using the Cauchy criterion it is enough to show that for each $\epsilon>0$ there is an $N\in\Bbb N$ such that
$$\left|\sum_{k=m}^\infty(-1)^kx^{2k}\right|<\epsilon$$ for all $m\ge N$ (for the chosen $x\in(-1,1)$). Then it is enough to see that $$\left|\sum_{k=m}^\infty(-1)^kx^{2k}\right|< |x|^{2m}$$
A: This converges to
$$f(x) = \frac{1}{1+x^2}$$
Thus $$|f(x) - \sum_{n=0}^Nf(x_n)|^2 = \frac{(x^2)^{4N+4}}{(1+x^2)^2}$$
Integrating this
$$\int_{-1}^1 \frac{(x^2)^{4N+4}}{(1+x^2)^2} dx \leq \int_{-1}^1(x^2)^{4N+4} dx = \frac{2}{4N+5}$$
$$\lim_{N\to \infty} \frac{2}{4N+5} = 0$$
Therefore, the sum converges in $L^2$ sense in this interval
