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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

$$lim_{Nto infty}int_a^b |f(x) - sum_{n=0}^Nf(x_n)|^2 dx = 0$$

but this problem has no target function

$sum_{n=0}^{N}(-1)^{n} x^{2n}=frac {1-(-x^{2})^{N+1}} {1+x^{2}}$. 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 $L^{2}(-1,1)$.

HINT: you dont need a target function. Using the Cauchy criterion it is enough to show that for each $epsilon>0$ there is an $NinBbb N$ such that

$$left|sum_{k=m}^infty(-1)^kx^{2k}right|<epsilon$$ for all $mge N$ (for the chosen $xin(-1,1)$). Then it is enough to see that $$left|sum_{k=m}^infty(-1)^kx^{2k}right|< |x|^{2m}$$

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

$$int_{-1}^1 frac{(x^2)^{4N+4}}{(1+x^2)^2} dx leq int_{-1}^1(x^2)^{4N+4} dx = frac{2}{4N+5}$$

$$lim_{Nto infty} frac{2}{4N+5} = 0$$

Therefore, the sum converges in $L^2$ sense in this interval

Hint. There is a target function $fin L^2(-1,1)$. For $xin(-1,1)$ we have a convergent geometric series:
$$f(x)=sum_{n=0}^infty (-1)^n x^{2n}=frac{1}{1+x^2}.$$