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Demonstrate the following equality:

$\log_e (a+bz) = \log_e a+\sum_{n=1}^\infty (-1)^{n-1}b^{-n}a^nz^nn^{-1}$ with $|z|<|a/b|$ , $ab\not=0$

What I've done:

$f(z) = \log_e (a+bz) -\log_e a +\log_e a = \log_e (1-(-ba^{-1}z))+\log_e a$

Integrating the geometric series:

$\log_e (1-x) =-\sum_{n=0}^\infty x^{n+1}(n+1)^{-1}$

Let $x=-ba^{-1}z$ and adding $\log_e a$ to both sides of the above equation:

$\log_e (1-(-ba^{-1}z)) + \log_e a=-[\sum_{n=0}^\infty ((-1)^{n+1}(ba^{-1}z)^{n+1}(n+1)^{-1}]+\log_e a$

Now... Am I forgetting something? Is this correct? If not, what went wrong?

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  • $\begingroup$ Should the summand be $(-1)^{n-1}b^na^{-n}z^nn^{-1}$? $\endgroup$ Dec 28, 2016 at 20:37
  • $\begingroup$ Hello. Not really, the solution is the above one, at least it is the one that the professor wrote on the sheet paper. But maybe, it was wrong all along... $\endgroup$
    – João Mota
    Dec 28, 2016 at 20:41
  • $\begingroup$ Nothing is wrong. Make the summation from $1$ (instead of $0$), and see how all $n+1$ turn to $n$. Then move $-$ inside the summation and see how the power of $-1$ changes. $\endgroup$
    – user58697
    Dec 28, 2016 at 20:57

1 Answer 1

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Could make it simpler:

\begin{align*} \ln (a+bz) &= \ln \left[ a\left(1+\frac{bz}{a} \right) \right] \\ &= \ln a+\ln \left(1+\frac{bz}{a} \right) \\ &= \ln a+\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \left( \frac{bz}{a} \right)^{n} \end{align*}

where $a>0$ and $\left| \dfrac{bz}{a} \right|<1$

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