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/sci/ - Science & Math

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5297882 No.5297882 [Reply] [Original]

Little Mathematical challenge for those of you interested:
Prove that "<span class="math">\displaystyle\sum\limits_{k=1}^{\infty}\left(\displaystyle\frac{1}{k \cdot x^k}\right)=\ln\left(\displaystyle\frac{x}{x-1}\right)[/spoiler] is true for any real "<span class="math">x[/spoiler]" greater than 1.

Only rule:
>Don't use Taylor series

Good luck and I will post my solution when a few of you post yours.

>> No.5297892

> hey sci help me do my homework
No thanks

>> No.5297895

> Hey guys my previous thread got deleted, or I deleted it cause someone posted it was homework
> So you wanna do my homework this time?

Still not, buddy.

>> No.5297914

>>5297882
>Its not homework guys I swear!

>> No.5297917

>>5297895
Actually, I deleted it because I didn't write the rule, which is "Don't use Taylor series".
I don't really give a fuck if you believe me or not. This is for people who want to have fun and/or waste time. If you don't want to do it, don't, it's that east.

>> No.5297945

Bump

>> No.5297978

>>5297892
>>5297895
>>5297914
Holy fucking shit /sci/ sucks cock. You cunts are constantly paranoid for no good reason and think every thing is either a troll or homework.
This was a genuine challenge for fun but it got ruined, so here is the solution just you can sleep assured that this wasn't another homework thread, faggots.

<span class="math">\displaystyle\sum\limits_{k=1}^{\infty}\left(\frac{1}{k \cdot x^k}\right)=\int \frac{d}{dx}\left(\sum\limits_{k=1}^{\infty}\left(\frac{1}{k \cdot x^k}\right)\right)dx=
\int \sum\limits_{k=1}^{\infty}\frac{d}{dx}\left(\frac{1}{k \cdot x^k}\right)dx=-\int \sum\limits_{k=1}^{\infty}\left(\frac{1}{x^{k+1}}\right)dx=-\int \left(\frac{1}{x-1}-\frac{1}{x}\right)dx=-\left(\ln(x-1)-\ln(x)\right)=\ln\left(\frac{x}{x-1}\right)[/spoiler]

>> No.5298003

>>5297978
fuck you OP I was almost there.
But it seems I was on the right path. anyway............. fuck you op

>> No.5298012

>>5298003
Why fuck me?
I'm sorry but I thought no body wanted to do it since the three first posts were all saying it was homework.

>> No.5298015

>>5297978
>highlight all TeX characters
>format switches to jsMath input so I can easily copy/paste it

Holy balls is that useful as fuck. I never even knew that until just now. Thanks, OP. This was a beautiful and useful discovery.

>> No.5298022

>>5297978
We have our reasons.

>> No.5298024

Which calc class was I supposed to learn how to integrate summations?

>> No.5298038

>>5298024
All you need to know to do this is the derivative of x^n and the integral of 1/x. No fancy stuff.
Also, you need to know that <span class="math">\displaystyle\sum\limits_{k=1}^\infty\left(\frac{1}{x^k}\right)=\frac{1}{x-1}[/spoiler]
>>5298022
Yeah, to ruin the fun for every body and be pessimistic about every thing in life.

>> No.5298046

>>5298038
That's where I got stuck actually, I was gonna post but you gave away the answer so...
how bout another one? I do wanna play and fuck these nazis

>> No.5298053

>>5298046
I also need to get back to my calc notes!!

>> No.5298061

>>5298046
I'm sorry, I don't have any challenges at the moment but I will post some in the future.
When I stumble on something interesting and I prove it, I usually post it on /sci/.
Sorry once again for ruining it for you.

>> No.5298077

>>5298061
lol don't worry, blame the aspies. you made me remember why math is fun

>> No.5298124

>>5298024
Probably the one where you first learned integration rules? Integral of (a+b) dx = integral of a dx + integral of b dx.

>> No.5298157

>>5298061
Grab a putnam problem. Harvard has one posted some where on their websites all the time.

>> No.5298240

>>5298157
This is much easier than a Putnam problem. Also, /sci/ never finished a Putnam problem.
Anyway, I found a new problem that I just proved and it's very delicious. I will post it in a few minutes and link to it from here.

>> No.5298258

Unrelated question here. I'm not sure if this is true for a general linear operator:

If A is not invertible, then AB is not invertible for any B. The case where B is invertible is trivial, but can't seem to make it for B not invertible.

>> No.5298395

>>5298258
Just look at any element not in the image of A