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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 $s.$

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

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If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ 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, $|x|<1$). $\endgroup$ – DanielWainfleet Mar 20 '16 at 12:55
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    $\begingroup$ I edited the question. $\endgroup$ – Raio Mar 20 '16 at 13:00
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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.

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