Check that $f(x)=\sum_{n=1}^{\infty}x^2(1-x^2)^{n-1}$ is continuous or not.

Question

Define $f:[0,1]\to \Bbb R$ by

$$f(x)=\sum_{n=1}^{\infty}x^2(1-x^2)^{n-1}$$

Check that $f$ is continuous or not.

Attempt: $$f(x)=\sum_{n=1}^{\infty}x^2(1-x^2)^{n-1}\\ =\lim\int_{0}^{1}x^2(1-x^2)^{n-1}$$ Now, Putting $x=\sin\theta$, then $\mathrm dx=\cos\theta\mathrm d\theta$, therefore integral reduces to

$$\lim\int_{0}^{\pi/2}\sin^2\theta(1-\sin^2\theta)^{n-1}\cos\theta\mathrm d\theta\\ =\lim\int_{0}^{\pi/2}\sin^2\theta\cos^n\theta\mathrm d\theta$$

Now from here the result will depend on $n$ i.e. $n=2m,n=2m+1$ In these two cases the result will be different, Hence $f$ is not continuous.

am I right? Different approaches are invited. Thank you.

Answer 1

To see that the series is geometric, hold x constant, compute $\frac {a_{n+1}} {a_n}=\frac {x^2(1-x^2)^n}{x^2(1-x^2)^{n-1}}=1-x^2$, which you can see is independant of $n$. At $x=0$, all your terms are trivially 0. At $x=1$, the first term is 1 (if you accept $0^0=1$ as the set theorists would have it!) and all the others are 0, so $f(1)=0$. Between them, $1-x^2<1$, so it converges. In between, you have the formula for the sum of a geometric series is $\frac {a_1}{1-r}=\frac {x^2}{1-(1-x^2)}=1$.

So, for $x\in (0,1)$, $f(x)=1$, so you have one jump discontinuity at 0

(editted as per comments)

Answer 2

The series converges pointwise to $f$ on $[0,1]$. Note that $f(0)=0$.

By the $M$-Test, $\sum_{n=1}^{\infty }x^2(1-x^2)^{n}$ converges uniformly on $[a,1];\ a\neq 0$ and so is continuous there.

But $\sum_{n=1}^{\infty }a^2(1-a^2)^{n}=a^2\sum_{n=1}^{\infty }(1-a^2)^{n}=\frac{a^{2}}{1-(1-a^{2})}=1$ and since this is true for all $0<a<1$ $f$ is not contiuous at $x=0$.