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I've looked through the Stack old questions, and searched the net, and I haven't found a proof that Cesaro summability implies Abel summability. Is the proof extremely complicated? Does anyone know a good reference?

I am also interested in the proof that if $\sum a_n$ and $\sum b_n$ are convergent, then the Cauchy product of the two sequences is Cesaro summable.

Gottfried Helms
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Eric Auld
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  • I'd try Hardy's "Divergent Series" for both questions – Chris Leary Oct 15 '13 at 17:57
  • @ChrisLeary Is the proof somewhat involved? – Eric Auld Oct 15 '13 at 18:04
  • Also Konrad Knopp's monography on series handles this question - and in it's german original it is even online at the göttingen digitizing center, however I don't know whether this is also true for the english version. The second question is explicitely answered in Chap 13, §276 and §277, theorem 8 and 9 (german edition) – Gottfried Helms Oct 15 '13 at 18:05
  • the first question is answered in §279, theorem 11, to the positive. All that don't look too much involved...(however I'm not at the moment inclined to reproduce or paraphrase the text here) All this is discussed under the bulletline of "Tauberian theorems", perhaps with this search key you're able to find focused resources online (wikipedia?) – Gottfried Helms Oct 15 '13 at 18:08
  • I think that this question (or, better, it's answer) also gives a proof. – tracing Jan 01 '15 at 03:32

1 Answers1

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Proof that Cesaro summable implies Abel summable : let $(a_n)_n$ be a sequence. Denote $$s^0_n = \sum_{n=0}^n a_n ~, \quad s^1_n = \sum_{n=0}^n s^0_n.$$ Suppose hat $\sum a_n$ is Cesaro summable, i.e. $\lim_n s^1_n/(n+1) = L$. Note that (easy calculation) $$\sum_{n=0}^\infty s^1_n x^n = (1-x)^{-1} \sum_{n=0}^\infty s^0_n x^n = (1-x)^{-2} \sum_{n=0}^\infty a_n x^n$$ (this calculation also prove that all the series converge for $|x|<1$). We have to prove that $\lim_{x \to 1} (1-x)^2 \sum_n s^1_n x^n = L$. Note that $\sum (n+1) x^n = (1-x)^{-2}$. Let $N>0$ and $x \in [0,1)$ and write $$(1-x)^2\sum_{n=0}^\infty s^1_n x^n = \sum_{n=0}^\infty (n+1)(1-x)^2x^n \times \frac{s^1_n}{n+1},$$ $$L = \sum_{n=0}^\infty (n+1)(1-x)^2x^n \times L.$$ Hence $$\left|(1-x)^2 \sum_{n=0}^\infty s^1_n x^n - L \right| \leq \sum_{n=0}^N (n+1)(1-x)^2x^n \times |\tfrac{s^1_n}{n+1}-L| + \sum_{n=N+1}^\infty (n+1)(1-x)^2x^n \times |\tfrac{s^1_n}{n+1}-L|.$$ From this formula is easy to conclude (the second term in the RHS is small for $N$ big indepandently from $x$, and the first term goes to $0$ for $N$ fixed and $x\to 1$)

Proof that $\sum (a*b)_n$ is Cesaro summable. Let $\sum a_n$ $\sum_n b_n$ be two series and let : $c_n = \sum_{p+q=n} a_p b_q$ and $X_N = \sum_{n=0}^N x_n$ (where $x=a,b,c$). I let you show that $$ \sum_{N=0}^M C_N = \sum_{P+Q=M} A_P B_Q.$$ (Hint : you can show for example that both sides are equal to $\sum_{p+q \leq M} (M+1-p-q)a_p b_q$). From this, it is easy to conclude that $\lim_M (M+1)^{-1} \sum_0^M C_N =A.B$ (where $A=\sum a_n = \lim_n A_n$, $B=\sum b_n = \lim_n B_n$), because if $M$ is big then $P$ and $Q$ are big hence, $A_P$ is closed to $A$ and $B_Q$ is closed to $B$.

Eric Auld
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user10676
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  • Thank you for your answer, sir. How can we see that $$\sum_{n=0}^\infty s^1_n x^n = (1-x)^{-1} \sum_{n=0}^\infty s^0_n x^n = (1-x)^{-2} \sum_{n=0}^\infty a_n x^n$$ – superAnnoyingUser Jun 25 '14 at 10:33
  • @Student It is called "Frobenius Lemma" ;) – Maman Apr 30 '16 at 21:40
  • @Maman I cannot find it under that name, could you give any reference? It's easy indeed, but maybe I'll find sth more general. Thanks – whereabouts Nov 14 '16 at 14:56
  • @whereabouts Makarov "selected problems" and Benedetto "spectral synthesis" for more explanations – Maman Nov 15 '16 at 15:21
  • @Maman could You maybe explain the above relation from superAnnoyingUser for which You wrote that it's called Frobenius Lemma? I tried to find Makarov "selected problems" and Benedetto "spectral synthesis" free pdf-s, as I understood that there I can find the proofs, but I cannot find a free pdf. Thanks! – syphracos Nov 28 '22 at 10:33
  • @syphracos we know that, for $|x| < 1$, $\sum_1^Na_nr^n = s_Nx^N - \sum_1^{N-1}(x^{n+1}-x^n)s_n$. You can check that by doing algebra. Rearranging again, we get $\sum_1^Na_nx^n = s_Nr^N + (1-x)\sum_1^{N-1}s_nx^n$. let $\rho_n := \frac{s_1 + ... + s_n}{n}$. Since $s_Nx^N = (N\rho_N - (N-1)\rho_{N-1})x^N$, the fact that Cesaro summability implies that $\rho_n$ are bounded and the fact that $\lim_{N \to \infty} Nx^N = 0$ together imply that $s_Nx^N \to 0$. Thus, taking the limit as $N \to \infty$ on $\sum_1^Na_nx^n = s_Nr^N + (1-x)\sum_1^{N-1}s_nx^n$ gives us $\sum a_nx^n = (1-x)\sum s_n x^n$. – Squirrel-Power Dec 24 '22 at 00:41
  • Can this proof possibly be extended to prove the statement: “$(C, \alpha)$ summability implies Abel summarily”? – Jono94 Mar 23 '24 at 08:23