I have read the proof for the Leibniz criterion. Underneath there was a remark that I did not understand it says that if we have a Leibniz series than one can obtain $s$ sometimes faster by calculating the (arithmetic) mean value and with index shift.
With Leibniz series I mean that $(a_n)$ is a monotonically decreasing zero-convergent sequence. And the Leibniz series $\sum_{n=0}^{\infty}(-1)^na_n$ converges. Let $s$ be the value it converges to
My Question why
$\frac 1 2a_0+\frac 1 2\sum_{n=0}^{\infty}(-1)^n(a_n-a_{n+1})$
also converges to $s$?
And if
$|s-\sum_{n=0}^k(-1)^na_n|\leq a_{k+1}$
Why is
$|s-\big(\frac 1 2a_0+\frac 1 2\sum_{n=0}^{\infty}(-1)^n(a_n-a_{n+1})\big)|\leq a_{k+1} a_{k+2}$
Thanks for helping
What I have thought so far is that the if we have
$\frac 1 2a_0+\frac 1 2\sum_{n=0}^{\infty}(-1)^n(a_n-a_{n+1})$
then the Right Hand side cannot converge to $s$ it must somehow compensate the influence of the Always constant $a_0/2$ in a way that it converges to $s$. Maybe one can do someting with the distributive law but I am not sure if I am allowed to use it.
In particular if $L=\sum (-1)^n \frac{1}{n+1}$
then also
$L=\frac{1}{2}+\frac{1}{2}\big(\frac{1}{1\cdot 2}-\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}-\frac{1}{4\cdot 5}+ …\big)$
Best Answer
Generalizing on my comment: $$ \begin{align} \frac12a_0+\frac12\sum_{k=0}^n(-1)^k(a_k-a_{k+1}) &=\frac12a_0+\frac12\sum_{k=0}^n(-1)^ka_k-\frac12\sum_{k=0}^n(-1)^ka_{k+1}\\ &=\frac12a_0+\frac12\sum_{k=0}^n(-1)^ka_k-\frac12\sum_{k=1}^{n+1}(-1)^{k-1}a_k\\ &=\frac12a_0+\frac12\sum_{k=0}^n(-1)^ka_k+\frac12\sum_{k=1}^{n+1}(-1)^ka_k\\ &=\sum_{k=0}^n(-1)^ka_k+\frac12(-1)^{n+1}\,a_{n+1} \end{align} $$ If you are uncertain what you can use in an infinite sum, apply rules to a finite sum and then take the limit.