Euler's summability method for series convergence I was going through the book on Functional Analysis by Erwin Kreyzig, and I came across this as one of the exercises. As an application of functionals to summability of sequences, the following theorem has been established in the book:

Theorem: If $A = \left[ \alpha_{nk} \right]_{n, k \in \mathbb{N}}$ is an infinite real matrix and $\left( x_n \right)_{n \in \mathbb{N}}$ is a real sequence, then the summability method which takes $\left( x_n \right)_{n \in \mathbb{N}}$ to the real sequence $\left( y_n \right)_{n \in \mathbb{N}}$, given by
$$y_n = \sum\limits_{k = 0}^{\infty} \alpha_{nk} x_k,$$
is regular if and only if the following holds:
(a) $\lim\limits_{n \to \infty} \alpha_{nk} = 0$, for all $k \in \mathbb{N}$. 
(b) $\lim\limits_{n \to \infty} \sum\limits_{k = 1}^{\infty} \alpha_{nk} = 1$. 
(c) There is some $\gamma > 0$ such that for all $n \in \mathbb{N}$, we have $\sum\limits_{k = 1}^{\infty} \left| \alpha_{nk} \right| \leq \gamma$.

The exercise which I was talking about, transforms a given series $\sum\limits_{j = 0}^{\infty} \left( -1 \right)^j a_j$ to the series $\sum\limits_{n = 0}^{\infty} \dfrac{\Delta^n a_0}{2^{n + 1}}$, where $\Delta^0 a_j = a_j$ and $\Delta^n a_j = \Delta^{n - 1} a_j - \Delta^{n - 1} a_{j + 1}$ for all $j \in \mathbb{N} \cup \left\lbrace 0 \right\rbrace$.
I want to prove that this summability method is regular. So, I thought of using the above theorem. So, if we consider
$$x_n = \sum\limits_{j = 0}^{n} \left( -1 \right)^j a_j,$$
and
$$y_n = \sum\limits_{k = 0}^{n} \dfrac{\Delta^k a_0}{2^{k + 1}},$$
we can do some algebraic maipulations to obtain,
$$y_n = \sum\limits_{k = 0}^{n} \dfrac{1}{2^{k + 1}} \sum\limits_{j = 0}^{k} \binom{k}{j} \left( -1 \right)^j a_j.$$
Now, this is almost of the form required in the theorem. However, I am unable to manipulate the binomial coefficient outside the inner sum. Any hints to do so?

Edit: A summability method is called regular if it does not change the limits of convergent sequences.
 A: You need $y_n$ in terms of $x_k$ (not $a_k$!) with $0\leqslant k\leqslant n$. We're going to show that $$y_n=\frac{1}{2^{n+1}}\sum_{k=0}^n\binom{n+1}{k+1}x_k;$$ having this, the conditions (a), (b), (c) of the theorem are very easy to check.
Recall that $x_n=\sum\limits_{k=0}^n(-1)^k a_k$. Introducing $x_{-1}:=0$, we have $(-1)^ja_j=x_j-x_{j-1}$, and $$\Delta^k a_0=\sum_{j=0}^k\binom{k}{j}(x_j-x_{j-1})=\sum_{j=0}^k\left[\binom{k}{j}-\binom{k}{j+1}\right]x_j,$$ so that, doing the interchange $\sum\limits_{k=0}^n\sum\limits_{j=0}^k=\sum\limits_{j=0}^n\sum\limits_{k=j}^n$, we get $$y_n=\sum_{j=0}^n\alpha_{nj}x_j,\quad\alpha_{nj}:=\sum_{k=j}^n\frac{1}{2^{k+1}}\left[\binom{k}{j}-\binom{k}{j+1}\right].$$
Finally, $\alpha_{nj}=\frac{1}{2^{n+1}}\binom{n+1}{j+1}$ may be shown using induction on $n$. It clearly holds at $n=j$ (both formulae give $\alpha_{jj}=2^{-j-1}$), and the induction step $(n-1)\to n$ for $n>j$ goes like this: $$\alpha_{nj}=\underbrace{\frac{1}{2^n}\binom{n}{j+1}}_{=\alpha_{(n-1)j}}+\frac{1}{2^{n+1}}\left[\binom{n}{j}-\binom{n}{j+1}\right]=\frac{1}{2^{n+1}}\left[\binom{n}{j}+\binom{n}{j+1}\right]=\frac{1}{2^{n+1}}\binom{n+1}{j+1}.$$
