# Abel summability of arithmetic means of a sequence [closed]

Let $\displaystyle\lim_{x\to1^-}(1-x)\sum_{n=0}^{\infty}s_nx^n=s$ for $|x|<1$, i.e. the sequence $(s_n)$ be Abel summable to $s.$ How to prove the sequence of arithmetic means $\displaystyle (t_n)=\left(\frac{1}{n+1}\sum_{k=0}^{n}s_k\right)$ is also Abel summable to $s.$

## closed as off-topic by Math1000, Claude Leibovici, user296602, colormegone, RolandMar 20 '16 at 21:06

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Math1000, Claude Leibovici, Community, colormegone, Roland
If this question can be reworded to fit the rules in the help center, please edit the question.

• If $x$ can approach $1$ from any direction in the complex plane then $\sum_{n=0}^{\infty}s_n x^n$ exists for some $x$ with $|x|>1,$ which means the radius of convergence of $\sum_n s_n x^n$ exceeds $1,$ which implies $\sum_{n=0}^\infty s_k$ exists , and the Q becomes trivial. You should amend your Q to retsrict the values of $x$ (that is, $|x|<1$). – DanielWainfleet Mar 20 '16 at 12:55
• I edited the question. – Raio Mar 20 '16 at 13:00

Let $f(x)$ and $g(x)$ by

$$f(x) = \sum_{n \geq 0} s_n x^n \quad \text{and} \quad g(x) = \sum_{n \geq 0} t_n x^n.$$

Step 1. We claim that $g(x)$ converges absolutely for $|x| < 1$. For each $|x| < 1$, choose $r$ such that $|x| < r < 1$. Since $f$ has radius of convergence $\geq 1$, we know that $\sum_n |s_n| r^n =: M < \infty$. Then by the estimate

\begin{align*} |t_n x^n| & \leq \frac{|s_0| + \cdots + |s_n|}{n+1}|x|^n \\ &= \frac{|s_0|r^n + \cdots + |s_n|r^n}{n+1} \left( \frac{|x|}{r} \right)^n \leq \frac{M}{n+1} \left( \frac{|x|}{r} \right)^n, \end{align*}

$g(x)$ converges absolutely.

Step 2. By Step 1, we know that on the interval $(-1, 1)$ we can differentiate $g(x)$ term by term. Then it follows that

\begin{align*} (1 - x)\frac{d}{dx} xg(x) &= \sum_{n \geq 0} (s_0 + \cdots + s_n) (1 - x)x^n \\ &= \sum_{n \geq 0} (s_0 + \cdots + s_n) x^n - \sum_{n \geq 1} (s_0 + \cdots + s_{n-1}) x^n \\ &= f(x). \end{align*}

This means that $(xg(x))' = (1-x)^{-1}f(x)$. Therefore, by the L'hospital's rule, we obtain

$$\lim_{x \to 1^-} (1-x)g(x) = \lim_{x \to 1^-} \frac{xg(x)}{(1-x)^{-1}} = \lim_{x \to 1^-} \frac{(xg(x))'}{(1-x)^{-2}} = \lim_{x \to 1^-} (1-x)f(x) = s.$$

Remark. Here, it is not hard to check that the L'hospital's rule for so called $\infty / \infty$ form only requires that the denominator diverges to infinity. So we do not need to check whether $xg(x)$ diverges or not. One may avoid the use of L'hospital's rule by writing

$$(1-x)g(x) = \frac{1-x}{x} \int_{0}^{x} \frac{f(t)}{1-t} \, dt$$

and aplying some approximation-to-the-identity argument.