Differentiability of a power series.

power seriesreal-analysissequences-and-series

I am working on the following exercise :

Let $f : x \mapsto \displaystyle\sum\limits_{n=1}^{+\infty} \dfrac{(-1)^nx^{2n+1}}{4n^2-1}$. The question asks to study the differentiability of $f$ on its domain.

I found that $f$ is defined on $[-1,1]$, continuous on its domain (because the series is normally convergent), and it is immediately differentiable on $(-1,1)$ as a power series. However, I have some difficulty to prove that it is differentiable at $x=\pm 1$.

Here is what I tried, e.g. for $x=1$. I wrote that $$\dfrac{f(x)-f(1)}{x-1} =\sum\limits_{n=1}^{+\infty} \dfrac{(-1)^n}{4n^2-1}\left( \dfrac{x^{2n+1}-1}{x-1}\right) =\sum\limits_{n=1}^{+\infty} \dfrac{(-1)^n}{4n^2-1}\sum_{k=0}^{2n} x^k$$

If I can prove that this series is uniformly convergent on $[0,1]$, then it will give the existence on the limit when $x$ tends to $1$. However, I cannot prove that this series is uniformly convergent : I think that it satisfies Leibniz criterium for alternating series, but I am not sure on how to prove that the absolute value of its general term is decreasing w.r.t. $n$.

Is it possible to prove the uniform convergence of this series using Leibniz criterium ?

Or is there another way to prove that $f$ is/isn't differentiable at $x=1$ ?

Thanks a lot for any help !

EDIT : I just found a solution : differentiating $f$ on $(-1,1)$, one gets that $f'(x)=-x\arctan(x)$, and then, one can deduce that $$f(x)=\dfrac{1}{2}(-(x^2+1)\arctan(x)+x)$$

Since $f$ is continuous, then this formula is still true on $[-1,1]$, and hence, $f$ is differentiable at $x=\pm 1$.

However, my question still stands : is it possible to prove that $f$ is differentiable at $x=\pm 1$ without computing explicitely the sum ?

Best Answer

A possible solution is to use the following theorem:

Let $(f_n)$ be a sequence of differentiable functions on $[a,b]$ such that $(f_n')$ converges uniformly (on $[a,b]$) and, for some $c\in[a,b]$, the sequence of numbers $\big(f_n(c)\big)$ converges. Then $(f_n)$ converges uniformly to a function $f$ differentiable on $[a,b]$, and $(f_n')$ converges to $f'$.

... and Abel's theorem: if $s=\sum_n a_n$ converges, then $g(x)=\sum_n a_n x^n$ converges uniformly on $x\in[0,1]$ (and $\lim_{x\to1}g(x)=s$); incidentally, the latter is not stated in this form on the Wiki page, but seen elsewhere, and is easy to verify following the proof given there.

In our case, $f_n$ are partial sums of the given series, and $f_n'$ are partial sums of a power series that fits Abel's theorem.

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