Find $\log_e3 - \frac{\log_e9}{2^2} + \frac{\log_e27}{3^2} - \frac{\log_e81}{4^2} + ...$ 
Find $\log_e3 - \dfrac{\log_e9}{2^2} + \dfrac{\log_e27}{3^2} - \dfrac{\log_e81}{4^2} + ...$

What I Tried:- This is the same as :-
$$\ln3 - \frac{\ln3}{2} + \frac{\ln3}{3} - \frac{\ln3}{4} + \dots$$
$$\rightarrow \ln3\bigg(1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots\bigg) $$
$$\rightarrow \ln3\Bigg(\sum_{n=1}^{\infty} (-1)^{n+1}\frac{1}{n}\Bigg)$$
Now, this summation's value came to be $\log 2$, which I came to know after using Wolfram Alpha, but I am not sure how this came. Can someone help me?
Also, the answer should be $(\ln 3)(\log 2)$.
$\rightarrow (\ln 3)\bigg(\dfrac{\ln 2}{\ln 10}\bigg).$
But I have the options, in my question, as :-
$(a) (\log_e2)(\log_e3)$ 
$(b) (\log_e2)$ 
$(c) (\log_e3)$ 
$(d) \dfrac{\log_e5}{\log_3}$
So none of the options match. Can someone confirm?
Also, can someone show me how the summation results to $\log 2$? Thank You.
 A: Assuming that $\log_e=\ln$, the answer is (a) as you already indicated. Perhaps you know the Taylor series of $\log(1+x)=x-x^2/2+x^3/3\mp\ldots$. Setting $x=1$ yields the result.
A: Your arguments are correct. You end up with the result $\ln 3\cdot \ln 2$, which corresponds to answer (a).
Also, the reason why $1-\frac 12+\frac 13-\frac 14+\cdots=\ln 2$ is because
$$\ln x=\sum_{k=1}^{\infty} \frac{(-1)^{k-1}(x-1)^{k}}{k}$$
for $0<x\leq 2$. Putting $x=2$ yields the result.
A: *

*The summation value is $\ln 2$ Some people use $\log$ to refer to base $e$ instead of base $10$.


*This series is called the alternating harmonic series. It's usually the example used to show while a series is divergent (the harmonic series is divergent), the alternating version can be  convergent.


*The way to evaluate alternate harmonic is:
Hmmm...see here. But I recall a heuristic is kinda to do with the idea like
$$\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n$$
Then replace $x$ with $-x$ to get
$$\frac{1}{1+x} = \sum_{n=0}^{\infty} (-x)^n$$
And then integrate to get
$$\ln(1+x) = \sum_{n=0}^{\infty} \frac{(-x)^{n+1}}{n+1}$$
The problem is the whole radius/integral of convergence thing. I can't quite just plug in $x=1$ in the last part given that the original interval of convergence (for $\frac{1}{1-x}$) is $(-1,1)$. I recall there's a thing here like they all have the same radius (and centre) of convergence of $1$ (and centre of $0$) but then for the interval of convergence you check each of the end points for each series. (But maybe there's some shortcut for the interval of convergence.) Anyway just see the link above.
