Zany

I've read that before posting on here, that's why I thought the answer would then just be (ln (2))^2? but how, if that is even correct?

Identities? hmm, well I'm pretty sleepy, I'll think on this more tomorrow.

What I showed is exactly like the problem that was given to me on a problem printed on a paper by the professor. I have it infront of me, and everything is exact.

This is for a University class.

I have nothing to give, besides what I've attempted, by distributing in the outside to each part inside that parenthesis. And this is for a mathematical problem solving class. It was randomly giving, no other information was given.

That's the thing, I'm not really clear on how to look at it either... not sure if that's a good way to look at it (replacing m for n when you can just use n either way?)?... but the exact problem that was given is in this exact form: [math] \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} (1-\frac{1}{2}+\frac{1}{3}-...+\frac{(-1)^{n+1}}{n}) [/math]

I am pretty much trying to find the summation of this: [math] \sum_{n=1}^{\infty}( \frac{(-1)^{n-1}}{n}) (1-\frac{1}{2}+\frac{1}{3}-...+\frac{(-1)^{n+1}}{n}) [/math] Pretty much the summation of both those product... If that cleared up anything... So yes your interpretation is correct, sorry if it was unclear... the other stuff was just my attempt at going somewhere with the problem.

Find the summation of: [math] \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} (1-\frac{1}{2}+\frac{1}{3}-...+\frac{(-1)^{n+1}}{n}) [/math] We know this is an alternating harmonic series if we look at the first part is just [math] \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n} [/math] and multiplied to the second part which is [math] \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n} [/math], this is also another alternating harmonic series. So the product of these two should converge to ln(2) for each part right? But I don't know how to show how they converge... so the summation is just (ln(2))^2? Or another way I looked at it is, distributing... [math] \sum_{n=1}^{\infty} (1\frac{(-1)^{n-1}}{n}-\frac{1}{2}\frac{(-1)^{n-1}}{n}+\frac{1}{3}\frac{(-1)^{n-1}}{n}-...+\frac{(-1)^{n+1}}{n}\frac{(-1)^{n-1}}{n}) [/math] and then we can compute each summation separately, but yet, I still cannot get anywhere with that....