# Is this approach correct? I can't arrange my result, so that it's equal to the solution, is there any mistake? Power series.

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?

• Should the summand be $(-1)^{n-1}b^na^{-n}z^nn^{-1}$? Dec 28, 2016 at 20:37
• 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... Dec 28, 2016 at 20:41
• 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. Dec 28, 2016 at 20:57

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