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 [tex]-1<x\leq +1[/tex],)

consider the change of variables given by

where q is a constant, and for which

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

re-index to start with k=0 to get

where the order of summation was reversed according to the rule

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

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

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

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

[tex]f(x)=\sum_{k=1}^{\infty}(-1)^{k-1}a_{k}x^{k} [/tex]

consider the change of variables given by

[tex]x=\frac{y}{1-qy},[/tex]

where q is a constant, and for which

[tex]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) [/tex]

[tex]= \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} [/tex]

[tex]= \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} [/tex]

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

[tex] 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} [/tex]

re-index to start with k=0 to get

[tex] 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}}[/tex]

where the order of summation was reversed according to the rule

[tex]\sum_{k=0}^{\infty} \sum_{n=k}^{\infty} b_{k,n}= \sum_{n=0}^{\infty} \sum_{k=0}^{n}b_{k,n} [/tex]

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

[tex]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}}[/tex]

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

[tex]\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} }[/tex]

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

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