In the book Finite Markov Chains by John G.Kemeny and J.Laurie Snell, the authors introduce the following concept of Euler-summability of a sequence:

Having a sequence $(s_n)_{n \geq 0},$ we introduce its sequence of averages $(u_n)_{n \geq 0},$ where for each $n \geq 0,$ $u_n = \sum \limits_{i=0}^n {n \choose i} k^{n-i} (1-k)^i s_i$ for some number $k$ such that $0 < k < 1.$ If the sequence $(u_n)_{n \geq 0}$ converges to a limit $u,$ then we say that sequence $(s_n)_{n \geq 0}$ is Euler-summable to $u$*.

Then the authors mention, that if the sequence $(s_n)_{n \geq 0}$ converges to a limit $s$, then it's also Euler-summable to $s,$ i.e. its sequence of averages $(u_n)_{n \geq 0}$ converges to $s$ as well.

I was trying to prove it using the $\varepsilon$-approach, similar to the one used in Convergence of series implies convergence of Cesaro Mean, but I failed. If someone could give me any hint on how to do it or simply proved it, I would much appreciate it.

  • 1
    $\begingroup$ The sum of the weights is $1$, so you can assume $s_n \to 0$ (replace $s_n$ with $s_n - s$). Pick $\varepsilon > 0$ and $N$ such that $\lvert s_n\rvert < \varepsilon$ for $n \geqslant N$. Split the sum at $N$. The thing to see is that $$\sum_{i = 0}^N \binom{n}{i} k^{n-i}(1-k)^i \xrightarrow{n\to\infty} 0.$$ $\endgroup$ – Daniel Fischer Sep 8 '15 at 20:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.