# Let $c_n\ge0$ be a sequence. Sufficient conditions of $(c_n)$ such that $\lim 1/n\sum\limits_{k=1}^n c_k z^k=0$ for $|z|=1,z\ne 1$

Let $$(c_n)$$ be a sequence of non-negative reals which is bounded below and above i.e. $$m\le c_n\le M$$ for some $$m,M>0$$. But this is not enough to say about the limit of $$\frac{1}{n}\sum\limits_{k=1}^n c_kz^k$$ to be zero (this post). Suppose I assume $$\lim c_n$$ exists or $$(c_n)$$ is Cesaro summable, can we conclude anything? Observe that $$(z^n)$$ is Cesaro summmable with cesaro sum is equal to $$0$$. Can we say something about the Cesaro summability of the product sequence $$(c_nz^n)$$? This post shows that the product of Cesaro summable sequence may not be Cesaro summable.

Can anyone help me with some idea, suggestions? Thanks for your help in advance.

We need the sequence $$(|c_{k+1}-c_k|)_k$$ to have $$0$$ Cesaro sum. More specifically,

Let $$S_n=\sum_{k=1}^nc_kz^k$$ where $$c_k\in\mathbb C$$ and $$\frac1n\sum_{k=1}^n|c_{k+1}-c_k|\to0 \tag{\star},$$ then $$\frac1nS_n\to0$$ for all $$|z|\leq1$$ with $$z\neq1$$.

First, some remarks:

• $$(c_k)$$ being Cesaro summable is not enough, consider $$(c_k)=(1,2,1,2,...)$$ and $$z=-1$$.

• $$(c_k)$$ is convergent$$\;\Rightarrow(\star)$$, but the reverse is false, consider $$c_k=\sqrt k$$.

• $$(\star) \nRightarrow (c_k)$$ is Cesaro summable or bounded , consider $$c_k=\sqrt k$$.

• $$|c_{k+1}-c_k|\to0\Rightarrow(\star)$$, but reverse is false, consider $$c_k=1$$ if $$k$$ is a square and $$c_k=2$$ otherwise.

• $$(c_k)$$ is Cesaro summable and $$\big|\frac{c_{k+1}}{c_k}\big|\to1\Rightarrow(\star)$$, but reverse is false, consider the same example in the above line.

Proof sketch:

Let $$S_n(z)=\sum_{k=1}^nc_kz^k$$ and it follows \begin{align*} S_n(z)&=c_nz^n+c_{n-1}z^{n-1}+\cdots+c_1z\\[.6em] &=z^n(c_n-c_{n-1})+\big(z^n+z^{n-1}\big)(c_{n-1}-c_{n-2})+\cdots+\big(z^n+z^{n-1}+\cdots+z^2\big)(c_2-c_1)\\ &\ \ \ \ +\big(z^n+\cdots+z\big)c_1\\[.5em] &=z^n(c_n-c_{n-1})+\frac{1-z^2}{1-z}z^{n-1}(c_{n-1}-c_{n-2})+\cdots+\frac{1-z^{n-1}}{1-z}z^2(c_2-c_1)+\frac{1-z^n}{1-z}zc_1. \end{align*} Since $$|z|\leq1$$ we have $$|1-z^k|\leq2$$ for all $$k$$ and therefore $$\frac1n|S_n(z)|\leq\frac2{1-z}\left(\frac{c_1}n+\frac1n\sum_{k=1}^{n-1}|c_{k+1}-c_k|\right).$$ As $$n\to\infty$$, the right hand side tends to $$0$$ due to $$z\neq1$$ and the condition $$(|c_{k+1}-c_k|)$$ has $$0$$ Cesaro sum.

For a sanity check, note that without $$(c_k)$$ we have $$\frac1n\sum z^k\to0$$ since the 'directions' of all $$z^k$$ cancel each other out. With the weight $$(c_n)$$, the directions may not cancel out since some directions can have larger weights than others. Hence, to ensure $$\frac1n\sum c_kz^k\to0$$ we need $$(c_k)$$ to be sufficiently 'uniform', in the sense that the weights for all directions are roughly the same in average. This loosely justifies the condition $$|c_{k+1}-c_k|$$ having $$0$$ Cesaro sum.

Also, boundedness of $$c_k$$ seems rather irrelevant here, since $$c_k$$ being bounded doesn't tell us anything about whether $$c_k$$ are biased towards some directions. (Unless you want to control the magnitude of the weights)