# Need explanation , infinite sum

#### idontknow

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

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#### SDK

after reading this http://web.math.ucsb.edu/~cmart07/Evaluating Series.pdf , I tried a similiar example below.
In which part is the mistake ?

$$\displaystyle z=\ln\sum_{n=0}^{\infty} \frac{z^n }{n!}$$
Its already off the rails at this point. Its well known that the sum is the exponential function but as far as I can tell you are trying to prove this by manipulating this series. But in this equality you have already presupposed this since the exponential function is the inverse of the natural logarithm. So any conclusion you arrive at will be true, but completely circular.

If I'm misunderstanding you then you should explain what exactly you are trying to do here.

idontknow