after reading this http://web.math.ucsb.edu/~cmart07/Evaluating Series.pdf , I tried a similiar example below.

In which part is the mistake ?

\(\displaystyle z=\ln\sum_{n=0}^{\infty} \frac{z^n }{n!} \; \) ; \(\displaystyle \; z\ln\sum_{n=0}^{\infty}\frac{z^n }{n!} =\ln^2 \sum_{n=0}^{\infty}\frac{z^n }{n!} \).

\(\displaystyle \frac{d}{dz} [z\ln\sum_{n=0}^{\infty}\frac{z^n }{n!}] =\frac{d}{dz}[\ln^2 \sum_{n=0}^{\infty}\frac{z^n }{n!}] \; \) ; \(\displaystyle \; \ln\sum_{n=0}^{\infty} \frac{z^n }{n!} +Se^{-z} = 2Se^{-z}\ln\sum_{n=0}^{\infty} \frac{z^n }{n!} \; \) , where \(\displaystyle S=z^{-1} \sum_{n=0}^{\infty} \frac{nz^n }{n!}\).

(*) \(\displaystyle \: z+Se^{-z}=2Sze^{-z} \;\) ; \(\displaystyle \; S=\frac{ze^z }{2z-1}\).

is

In which part is the mistake ?

\(\displaystyle z=\ln\sum_{n=0}^{\infty} \frac{z^n }{n!} \; \) ; \(\displaystyle \; z\ln\sum_{n=0}^{\infty}\frac{z^n }{n!} =\ln^2 \sum_{n=0}^{\infty}\frac{z^n }{n!} \).

\(\displaystyle \frac{d}{dz} [z\ln\sum_{n=0}^{\infty}\frac{z^n }{n!}] =\frac{d}{dz}[\ln^2 \sum_{n=0}^{\infty}\frac{z^n }{n!}] \; \) ; \(\displaystyle \; \ln\sum_{n=0}^{\infty} \frac{z^n }{n!} +Se^{-z} = 2Se^{-z}\ln\sum_{n=0}^{\infty} \frac{z^n }{n!} \; \) , where \(\displaystyle S=z^{-1} \sum_{n=0}^{\infty} \frac{nz^n }{n!}\).

(*) \(\displaystyle \: z+Se^{-z}=2Sze^{-z} \;\) ; \(\displaystyle \; S=\frac{ze^z }{2z-1}\).

is

**S**correct ? can we get more things from**(*)**?
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