# Hardy's power series. Cesàro convergence

Doing my analysis homework i have come across the following power series known as Hardy's power series $$\sum\limits_{k=0}^{\infty}a_kx^k=\sum\limits_{k=0}^{\infty}(-1)^kx^{2^k}\mbox{ for x}\in[0,1],$$

i.e, $$a_k=(-1)^n$$ if $$k=2^n$$ and $$a_k=0$$ otherwise. This exercise asks to prove that the series $$\sum a_n$$ doesn't converge Cesàro wise (i.e. the Cesàro means $$\sigma_n$$ diverge). It also asks if there is Abel convergence for $$\sum a_nx^n$$.

To prove divergence of $$\sigma_n$$, we are told to prove that $$$$\liminf\sigma_n=1/3\neq\limsup\sigma_n=2/3\mbox{ as }n\rightarrow\infty.$$$$

Once you manage to show that, the issue of Abel convergence is easy enough, since the partial sums can only be $$0$$ or $$1$$; and the Hardy-Littlewood theorem tells us that if there were Abel convergence then there would be Cesàro convergence, hence the power series can not be Abel convergent.

The problem is that i have not been able to prove the divergence of $$\sigma_n$$, even with the hint provided by the exercise. Can someone help me please???

If you just look at the partial sums, it is $$s_{2^k+\ell}=\sum_{n=1}^{2^k+\ell} a_n = \begin{cases} 0 & k\text{ odd}\\ 1 & k\text{ even} \end{cases}$$ where $$0\leq\ell<2^k$$. In other words, start with a 1 (I'm going to assume you start the series off at $$n=1$$ not $$n=0$$), then put 2 0's, then 4 1's, then 8 0's, ... each time doubling the streak of the last but flipping 0s and 1s. So if you stop just after the streak of 1s then the density of 1s would be almost 2/3 of the time (except the first 1 doesn't get a half zero before it, so at that point we get $$\frac23+o(1)$$). On the other hand, if you stop just after the streak of 0's then it is $$\frac13$$. Stopping anytime inbetween would give you something between 1/3 and 2/3, so these are the liminf and limsup of Cesaro sums.