I am confused about the Hardy-Littlewood tauberian theorem. If we apply it with the sequence $a_n$ whose first few terms are given by $1,-1,1,1,-1,-1,1,1,1,1,-1,-1,-1,-1...$ where we put $2^n$ $1$s followed by $2^n$ $-1$s going on indefinitely, increasing $n$.
At the end of a period of $-1$s, there have been equally as many $+1$s and $-1$s meaning that $\frac{1}{n}\sum_{k=1}^na_k=0$. At the end of a period of $1$s, about $\frac{1}{3}$ of the numbers that have appeared thus far has been this last streak of $1$s and since all of the previous terms average out to zero this means that $\frac{1}{n}\sum_{k=1}^na_k\approx\frac{1}{3}$. Overall, this means that this sequence has no defined Cesàro mean.
On the other hand, when examining $(1-x)\sum_{n=1}^{\infty}a_nx^n$ as $x\to1^-$ using python I have noticed that it seems to converge to a specific value that is $\approx.170$.
This would contradic the Hardy Littlewood tauberian theorem since an asymptotic relation $(1-x)\sum_{n=1}^{\infty}a_nx^n\sim C$ would imply the relation $\frac{1}{n}\sum_{k=1}^na_k\sim C$ which does not seem to exist.
This case is an example of a more general issue I have with the Hardy-Littlewood tauberian theorem. Using L'hopitals we can easily get that for bounded sequences $a_n$ we have
$$\lim_{x\to1^-}\left[\left(1-x\right)\sum_{n=0}^{\infty}a_{n}x^{n}-\left(1-x\right)^{2}\sum_{n=0}^{\infty}na_{n}x^{n-1}\right]=0$$
and so by differentiating $f(x)=(1-x)\sum_{n=1}^{\infty}a_nx^n$ we get that $f'(x)=o\left(\frac{1}{1-x}\right)$. If $\lim_{x\to1^-}f(x)$ doesn't converge then the difference between consecutive minima and maxima is $>O(1-x)$ and thus my the mean value theorem there is a point between the two with $f'(x)>O\left(\frac{1}{1-x}\right)$. Since there would be infinitely many point like this, we have a contradiction based off the growth rate of $f'(x)$.
This means that $f(x)$ converges as $x\to1^-$, but this implies that $\lim_{n\to\infty}\sum_{k=1}^n a_k$ converges by the Hardy Littlewood tauberian theorem which obviously is not true for a general bounded sequence $a_n$.
Can anyone explain to me what I am doing wrong?
This is more of a comment that it got too long; the post above is very sloppy and confusing with a bunch of mistaken statements:
for example "$f(x)=o\left(\frac{1}{1-x}\right)$" is trivial when $f(x)=(1-x)\sum{a_nx^n}, |a_n| \le C$ since $f(x)$ is bounded at $1$
(for $a_n$ real it is just trivial that for $0<x<1, -C/(1-x) \le \sum{a_nx^n} \le C/(1-x)$ and for complex $a_n$ separation in real and imaginary parts will do)
However, this doesn't imply in any way shape or form that $f(x) \to C, x \to 1, x<1$, while also the fact that an analytic function is bounded in the unit disc doesn't imply any growth condition for the derivative (there are infinite Blaschke products $|B(z)| <1, |z|<1, |B(e^{it})|=1$ a.e. s.t $B'$ is not $O(1/(1-z)^a)$ for any $a>0$ however large,
Similarly the last statement is wrong as the H-L Tauberian theorem implies only $a_n \to C$ or the Caesaro mean $(a_0+..a_n)/(n+1) \to C$ in the context of $a_n$ bounded
As for the example given, by a simple computation $g(x)=(1-x)\sum a_nx^n=\sum b_nx^n$ where $b_n$ is a $4/3$ gap sequence since $b_n=a_n-a_{n-1}, n \ge 1$ so $b_n \ne 0$ only when the $1$ transitions to $-1$ and viceversa, so roughly at $2^{n+1}, 3.2^n$ where $b_n = \pm 2$ (except for $b_0=1$) - hence the gaps are $3/2, 4/3$ respectively so overall $4/3$ (or maybe $4/3-\epsilon$) works
But then it is a classic theorem that a gap power series satisfies $g(x) \to C, x \to 1$ iff $\sum {b_n}=C$ (we need no Tauberian conditions in other words), theorem that follows easily from the result that if $|g(x)| \le M, 0 \le x <1$ then $|\sum_{k \le n} {a_k}| \le C_{\rho}M$ for all $n$ where $\rho >1$ is a gap for the coefficients of $g$, namely, $b_{n_k} \ne 0$ implies $n_{k+1} \ge \rho n_k$ (eventually). In particular since for the $g$ in the example, $b_n$ doesn't converge to zero, $g(x)$ cannot converge and it oscillates.
To try and find oscillating sequences, try $x_n=1-1/2^n, y_n=1-1/(3 \times 2^n)$ as $g(x_n), g(y_n)$ should converge to different values (one can do estimates and show that theoretically too, but would check computationally first).
So no contradiction with classical theory anywhere. As for mkistakes, the statement about the derivative is wrong and the computations done were not done at the right numbers or maybe the diffeerent convergence is too slow (though I think for $n$ about $20$ so for $x,y$ in the $1-1/10^6$ range those numbers should be different already when truncating at $n$ abut $8 \times 10^6$)