# Euler's Series Transformation

1. May 6, 2006

### benorin

I am working on my thesis, one topic therein is Euler's series transformation, would you kindly look over my derivation of it?

Workin with the function f(x) defined by the following power series (convergent for at least $$-1<x\leq +1$$,)

$$f(x)=\sum_{k=1}^{\infty}(-1)^{k-1}a_{k}x^{k}$$​

consider the change of variables given by

$$x=\frac{y}{1-qy},$$​

where q is a constant, and for which

$$x^k=\left( \frac{y}{1-qy}\right) ^k = y^k \left( \frac{1}{1-qy}\right) ^k = \frac{y^k}{(k-1)!q^{k-1}}\frac{d^{k-1}}{dy^{k-1}} \left( \frac{1}{1-qy}\right) = \frac{y^k}{(k-1)!q^{k-1}}\frac{d^{k-1}}{dy^{k-1}} \left( \sum_{n=0}^{\infty}q^ny^n \right)$$
$$= \frac{y^k}{(k-1)!q^{k-1}} \left( \sum_{n=k-1}^{\infty}\frac{n!}{(k-n+1)!}q^ny^{n-k+1} \right) = \frac{1}{q^{k-1}} \sum_{n=k-1}^{\infty} \left(\begin{array}{c}n\\k-1\end{array}\right) q^ny^{n+1}$$​

so that we have (applying the change of variables to f(x),)

$$f\left( \frac{y}{1-qy}\right) = \sum_{k=1}^{\infty}(-1)^{k-1}a_{k}\left( \frac{y}{1-qy}\right) ^{k} = \sum_{k=1}^{\infty}(-1)^{k-1}a_{k} \frac{1}{q^{k-1}} \sum_{n=k-1}^{\infty} \left(\begin{array}{c}n\\k-1\end{array}\right) q^ny^{n+1}$$​

$$f\left( \frac{y}{1-qy}\right) = \sum_{k=0}^{\infty} \sum_{n=k}^{\infty} (-1)^{k}\frac{a_{k+1}}{q^{k}} \left(\begin{array}{c}n\\k\end{array}\right) q^ny^{n+1} = \sum_{n=0}^{\infty} \sum_{k=0}^{n} (-1)^{k}\frac{a_{k+1}}{q^{k}} \left(\begin{array}{c}n\\k\end{array}\right) q^ny^{n+1} = \sum_{n=0}^{\infty} q^ny^{n+1} \sum_{k=0}^{n} (-1)^{k} \left(\begin{array}{c}n\\k\end{array}\right) \frac{a_{k+1}}{q^{k}}$$​

where the order of summation was reversed according to the rule

$$\sum_{k=0}^{\infty} \sum_{n=k}^{\infty} b_{k,n}= \sum_{n=0}^{\infty} \sum_{k=0}^{n}b_{k,n}$$​

Also note, that (according to the transformation variables given above,) the value $$y=\frac{1}{q+1}$$ corresponds to $$x=1$$; thus, considering f(1) gives

$$f(1) = \sum_{k=1}^{\infty}(-1)^{k-1}a_{k} = \frac{1}{q+1}\sum_{n=0}^{\infty} \left( \frac{q}{q+1}\right) ^{n} \sum_{k=0}^{n} (-1)^{k} \left(\begin{array}{c}n\\k\end{array}\right) \frac{a_{k+1}}{q^{k}}$$​

which, upon setting q=1, gives Euler's Series transfromation as a special case, namely

$$\boxed{ \sum_{k=1}^{\infty}(-1)^{k-1}a_{k} = \sum_{n=0}^{\infty} \frac{1}{2^{n+1}}\sum_{k=0}^{n} (-1)^{k} \left(\begin{array}{c}n\\k\end{array}\right) a_{k+1} }$$​

which is useful for series convergence acceleration (sometimes) and for the analytic continuation for functions defined by series. Any thoughts?

Last edited: May 6, 2006
2. May 6, 2006

### shmoe

Why the alternating series? This should work for any complex series, no? (given convergence of the power series).

Why is changing the order of summation justified here?

I didn't check every exponent or index carefully, but it looks essentially correct. The basic idea is simple enough. Have you looked at Hardy's "Divergent Series" book? It covers the basics, though I don't know if it's any more in depth than Knopp's.

3. May 6, 2006

### benorin

The alternating series is just for a slight notational convenience (using the forward difference operator $$\Delta$$ is nice this way, since $$\Delta ^{n} (a_k) = \sum_{k=0}^{n} (-1)^{k} \left(\begin{array}{c}n\\k\end{array}\right) a_{k+n}$$ ). I am not sure about the convergence issues on this one, the convergence is required on that segment of the real axis so that it is defined for x=+1 (which goes well with Abel's theorem, giving continuity at x=+1 if it converges there) and since it is a power series and so must converge with a radius of at least R=1; but note that the alternating series is just a construct to establish Euler's series transformation, and since I wish to use Euler's T-form primarily for analytic continuation, it is the coefficients of that alternating series (which are, in practice, functions of perhaps several complex variables) that matters most to me.

Last edited: May 6, 2006