Prove that:
If $\sum c_n$ is Abel summable to $s$ and $c_n=O(\frac{1}{n})$ , then $\sum c_n $ converges to $s$.
"A series of complex number $\sum_{n=0}^{\infty} c_n $ is said to be Abel summable to $s$ if for every $0 \le r <1$ ,the series $A(r)=\sum_{k=0}^{\infty} c_kr^k$ converges ,and $\lim_{r\rightarrow 1^-} A(r)=s$."
The following quote is from the book Jacob Korevaar: Tauberian Theory from Section 7.1: Hardy-Littlewood Tauberians for Abel Summability.
Littlewood [1911] answered Hardy's question whether the condition '$na_n \to 0$' in Tauber's theorem could be relaxed to boundedness of the sequence $\{na_n\}$.
Theorem 7.1. (Littlewood) $\sum a_n$ is Abel summable and $|na_n| < C$ $\Rightarrow$ $\sum a_n$ converges.
This 'big O-theorem' for Abel summability is more difficult than the earlier results. The theorems in this section have attracted much interest and invited many alternative proofs, frequently with Theorem 7.3 as the first step towards Theorem 7.1. Littlewood's original proof was rather complicated; his key tool was repeated differentiation; cf. Section 17. For comments on Littlewood's fundamental article of 1911, see his Collected Papers [1982]; for the history of his discovery, see Littlewood [1953], A first simple proof for the Theorem was found by Karamata [1930a]; see Section 11 for his method. A related more direct proof by Wielandt [1952] will be described in Section 12.
Theorem 7.3. (Hardy and Littlewood) One has the following implication: $\sum a_n$ is Abel summable and $s_n >-C$ $\Rightarrow$ $\sum a_n$ is Cesaro summable.
The following quote is from the book Boos: Classical and modern methods in summability
H. Tietz and K. Zeller drew my attention to a recent paper (cf. [240]) in which they give a modification of Wielandt's well-known elegant proof of the Hardy-Littlewood o-Tauberian theorems for the Abel method. This is an elementary proof and I decided to use the material of this paper for the important part of Section 4.4....
The shortest proofs are due to Wielandt [246] and Karamata [124] and, by modifying Wielandt's proof, to Tietz and Zeller [240]. We follow the lines of Tietz and Zeller's proof which avoids integrals and is based on the Weierstrass approximation theorem.
References mentioned above:
Perhaps you already know but this is not a "simple theorem". A proof with many other information can be found in K.Knopp's "Theory and Applications of Infinite series", Springer, pp.500-506
I can do better: here you find what you need.
Since $n c_{n}\to 0$, the sequence $(|n c_{n}|)$ is $(C,1)$ summable to $0$. i.e. $$ \lim_{n\to\infty}n c_{n}=0\tag{1}$$ Given $\epsilon >0$, it follows from our hypothesis and equation$(1)$ that there exists $N\in I$ such that $$|n c_{n}|<\frac{\epsilon}{3}, n\geq N\tag{2}$$ Thus $$\frac{1}{n}\sum_{k=1}^{n}|k c_{k}|<\frac{\epsilon}{3}, n\geq N\tag{3}$$ and such that $$|L-\sum_{k=0}^{\infty}c_{k} x^{k}|<\frac{\epsilon}{3} \qquad (\text{for }1-\frac{1}{N}<x<1)\tag{4}$$ For any $n\in N$ and $x\in(0,1)$ we have, \begin{equation*} \begin{split} L-\sum_{k=0}^{n}c_{k}&=L-\sum_{k=0}^{\infty}c_{k}x^{k}+\sum_{k=0}^{\infty}c_{k}x^{k}-\sum_{k=0}^{n} c_{k}\\ &= L-\sum_{k=0}^{\infty}c_{k}x^{k}+\sum_{k=1}^{n}c_{k}(x^{k}-1)+\sum_{k=n+1}^{\infty}c_{k}x^{k}. \end{split} \end{equation*} Hence \begin{equation*} \begin{split} |L-\sum_{k=0}^{n}c_{k}|&\leq |L-\sum_{k=0}^{\infty}c_{k}x^{k}|+\sum_{k=1}^{n}|c_{k}|(1-x^{k})+\sum_{k=n+1}^{\infty}|c_{k}|x^{k}\\ &=I_{1}+I_{2}+I_{3} \end{split} \end{equation*} For any $n\geq N$, choose $x$ such that $1-\frac{1}{n}<x<1-\frac{1}{n+1}$. Then $1-\frac{1}{N}\leq 1-\frac{1}{n}<x<1$ and so by $(4)$, $$I_{1}=|L-\sum_{k=0}^{\infty}c_{k}x^{k}|<\frac{\epsilon}{3}.$$ Now, $1-x^{k}=(1-x)(1+x+x^{2}+...+x^{k-1})\leq k(1-x)$, for any $k\in N$. Hence, since $1-x<\frac{1}{n}$, we have $$1-x^{k}\leq k(1-x)<\frac{k}{n}\tag{5}$$ By $(5)$ and $(3)$ we then have $$I_{2}=\sum_{k=1}^{n}|c_{k}|(1-x^{k})<\frac{1}{n}\sum_{k=1}^{n}|k c_{k}|<\frac{\epsilon}{3}$$ To estimate $I_{3}$ we have using $(2)$, \begin{equation*} \begin{split} I_{3}&=\sum_{k=n+1}^{\infty}|k c_{k}|\frac{x^{k}}{k}\\ &< \frac{\epsilon}{3}\sum_{k=n+1}^{\infty}\frac{x^{k}}{k}\\ &\leq \frac{\epsilon}{3(n+1)}\sum_{k=n+1}^{\infty}x^{k}\\ &\leq \frac{\epsilon}{3(n+1)}\sum_{k=0}^{\infty}x^{k}\\ &= \frac{\epsilon}{3(n+1)(1-x)}. \end{split} \end{equation*} But $x<1-\frac{1}{n+1}$ and so $1-x>\frac{1}{n+1}$. Thus $(n+1)(1-x)>1$ and so $$I_{3}\leq \frac{\epsilon}{3(n+1)(1-x)}<\frac{\epsilon}{3}.$$ From $I_{1}, I_{2}$ and $I_{3}$, it follows that, $$|\sum_{k=0}^{n}c_{k}-L|\leq I_{1}+I_{2}+I_{3}<\epsilon., n\geq N$$ Which proves that $\sum_{k=0}^{\infty}c_{k}$ converges to $L$.
