Help with Summation Problem

[Zany]

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....

[Schrödinger's hat]

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....

 

I'm a little bit unclear on what you're trying to work out.

 

This:

[math] \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} (1-\frac{1}{2}+\frac{1}{3}-...+\frac{(-1)^{n+1}}{n}) [/math]

 

I interpret as:

[math] \sum_{n=1}^{\infty} \left(\frac{(-1)^{n-1}}{n} (1-\frac{1}{2}+\frac{1}{3}-...+\frac{(-1)^{n+1}}{n}) \right)[/math]

 

But your inclusion of:

[math] \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n} [/math]

 

Seems to imply you've bracketed it as:

[math] \left(\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}\right) (1-\frac{1}{2}+\frac{1}{3}-...+\frac{(-1)^{n+1}}{n}) [/math]

With an implicit (independant) sum on the second part.

 

I also can't seem to find the exact link between either of these and your third statement.

 

Either you've done some intermediate steps that I've missed because my brain isn't entirely switched on right now, I'm otherwise mis-interpreting what you've said, or you've made a mistake of some kind.

[Zany]

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.

Edited November 27, 2011 by Zany

[Schrödinger's hat]

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.

 

Somewhat. Just to be clear that would be able to be rewritten as:

[math] \sum_{n=1}^{\infty}\left(\frac{(-1)^{n-1}}{n}\sum_{m=1}^n\frac{(-1)^{m+1}}{m}\right)[/math]

Correct?

[Zany]

Somewhat. Just to be clear that would be able to be rewritten as:

[math] \sum_{n=1}^{\infty}\left(\frac{(-1)^{n-1}}{n}\sum_{m=1}^n\frac{(-1)^{m+1}}{m}\right)[/math]

Correct?

 

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]

[Schrödinger's hat]

Well, now that we've established that, I have to remember how to deal with an infinite series. :/

Start off slow.

The bit in the bracket is going to be less than n.

The denominator of the fraction on the left is n.

The sign is alternating.

So it's an alternating series where the terms decrease, therefore it converges.

 

Not terribly useful, but it's a start.

We also know that it's somewhere in the vicinity of (ln(2))^2 as you correctly surmised. Because the term in the brackets is an ever-improving approximation of ln(2). But it won't be exactly.

 

Now I have to think back to first year to make any further progress. Excuse me while I retrieve it from the long term memory archives.

Loading...

 

I'm afraid I'm coming up somewhat empty.

Perhaps you could give me a little bit of context, it may provide a few clues.

What is your current mathematical level?

Where did you encounter this sum (was it an exercise in class? If so what type of class?)

From what you've said so far I would surmise it's a Calculus 101 course or similar, but the requisite trick for this one eludes me.

The n+1 vs n-1 looks like it was an intentional clue, but I see how it helps.

Edited November 27, 2011 by Schrödinger's hat

[Zany]

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.

[Schrödinger's hat]

(what level class? University? Highschool?)

[Zany]

This is for a University class.

[michel123456]

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]

 

I suspect an error in notation. Are you definitely sure that is the accurate question? Or is this a transcript from an oral statement?

 

I mean maybe it is this

[math] \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} (1-\frac{1}{2}-\frac{1}{3}-...+\frac{(-1)^{n+1}}{n}) [/math]

Edited November 27, 2011 by michel123456

[Zany]

I suspect an error in notation. Are you definitely sure that is the accurate question? Or is this a transcript from an oral statement?

 

I mean maybe it is this

[math] \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} (1-\frac{1}{2}-\frac{1}{3}-...+\frac{(-1)^{n+1}}{n}) [/math]

 

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.

Edited November 27, 2011 by Zany

[michel123456]

O.K. i said thast because i smell a trick.

what is the result of [math](-1)^{n+1} [/math]

 

and what is the result of [math]\frac{1}{2}+\frac{1}{3}+...[/math]

 

oops, I am wrong.

The trick is elsewhere.

I found it (partly), you'll find it (completely). There are identities.

Edited November 27, 2011 by michel123456

[Zany]

O.K. i said thast because i smell a trick.

what is the result of [math](-1)^{n+1} [/math]

 

and what is the result of [math]\frac{1}{2}+\frac{1}{3}+...[/math]

 

oops, I am wrong.

The trick is elsewhere.

I found it (partly), you'll find it (completely). There are identities.

 

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

[michel123456]

http://en.wikipedia.org/wiki/Harmonic_series_(mathematics)#Alternating_harmonic_series

[Zany]

http://en.wikipedia....es_(mathematics)#Alternating_harmonic_series

 

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?

[imatfaal]

Just a few thoughts

 

[math] \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}[/math] is identical to [math] \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n} [/math]

 

for all values of n - as [math] (-1)^{n+1}=(-1)^{n-1}[/math] for all n

 

additionally [math]( (-1)^{n+1})((-1)^{n-1}) = ( (-1)^{n-1})((-1)^{n-1}) = 1[/math] for all values of n

 

thus cannot you reduce the whole mess to the sum of [math] \frac{1}{n^2} [/math] which off the top of my head does not converge-

 

edit (i am not sure that you can )- in fact pretty sure you cannot and 1/n^2 does converge. apart from that....

Edited November 28, 2011 by imatfaal