Euler's Method and induction

Question

Euler's Method for series associates with a given series $\sum\limits_{j=0}^\infty(-1)^ja_j$ the transformed series $\sum\limits_{n=0}^\infty\frac{\Delta^n a_0}{2^{n+1}}$ where $\Delta^0a_j=a_j$, $\Delta^na_j=\Delta^{n-1}a_j-\Delta^{n-1}a_{j+1}$, $j=1,2,\cdots.$

I am given that $\ln 2 = 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+-\cdots = \frac{1}{1\cdot2}+ \frac{1}{2\cdot 2^2}+\frac{1}{3\cdot 2^3}+\frac{1}{4\cdot 2^4}\cdots$.

I need to prove that $ \sum\limits_{n=0}^\infty\frac{\Delta^n a_0}{2^{n+1}} = \frac{1}{2}+ \frac{1}{2\cdot 2^2}+\frac{1}{3\cdot 2^3}+\frac{1}{4\cdot 2^4}\cdots$, but I can not figure out how. I attempt to by induction and the base case is simple enough. However, when trying to use the induction hypothesis I get stuck.

If I assume that $\sum\limits_{n=0}^k\frac{\Delta^n a_0}{2^{n+1}} = \frac{1}{2}+ \frac{1}{2\cdot 2^2}+\frac{1}{3\cdot 2^3}+\frac{1}{4\cdot 2^4}\cdots\frac{1}{(k+1)\cdot 2^{k+1}}$ and I add $\frac{1}{(k+2)\cdot 2^{k+2}}$ to both sides. Then I just need to show that $\Delta^{k+1}a_0 = \frac{1}{k+2}$. However I don't see how I can.

So I know that $\Delta^{k+1}a_0 = \Delta^{k}a_0-\Delta^{k}a_1$. From the induction hypothesis I know that $\Delta^{k}a_0 = \frac{1}{k+1}$. But I would have to work with $\Delta^{k}a_1$ recursively. So I am not sure how to prove this.

Any input would be very welcome!

Answer 1

A formula for $ \Delta ^n a_j$ is given by \begin{eqnarray*} \Delta ^n a_j = \frac{n! \ j!}{(n+j+1)!} \end{eqnarray*} We now need to use double induction ... it suffice to show that the formula $\Delta^na_j=\Delta^{n-1}a_j-\Delta^{n-1}a_{j+1}$ holds, which is easy \begin{eqnarray*} \Delta ^n a_j = \Delta^{n-1}a_j-\Delta^{n-1}a_{j+1} =\frac{(n-1)! \ j!}{(n+j)!}-\frac{(n-1)! \ (j+1)!}{(n+j+1)!}= \frac{n! \ j!}{(n+j+1)!} \end{eqnarray*}