Sequences and Series/Power series: Difference between revisions

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{{proof|By Abelian partial summation, we have

1nxanzn=zxA(x)ln(z)1xA(t)ztdt

for |z|<1 and x1, where we denote as usual

A(x):=1nxan.

Substituting z=exp(w), we get

1nxanexp(wn)=exp(wx)A(x)w1xA(t)exp(wt)dt.

We then put

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