anonymous
  • anonymous
summation to infinity
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[\sum_{k=1}^{\infty} (-1)^k \frac{ 1 }{ k }\]
anonymous
  • anonymous
I don't know how to approach this. How would I start?
anonymous
  • anonymous
I could list out a bunch of terms and add them together in my calculator, but I don't like doing it that way.

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anonymous
  • anonymous
it's the easy way, not the best way
amistre64
  • amistre64
try to split the positive and negatives ... is an idea
anonymous
  • anonymous
you trying to find the actual sum?
anonymous
  • anonymous
yes
anonymous
  • anonymous
jeez i would first of all cheat then look for a power series representation of a function
amistre64
  • amistre64
\[1-\frac12x+\frac13x^2-\frac14x^3\pm...~:~x=1\]
amistre64
  • amistre64
maybe start with an x?
anonymous
  • anonymous
yeah exactly but first i cheated so i would know where i was headed
amistre64
  • amistre64
ive got classes today .... so i wont be back for about 2 hours. My advice is to post sooner :)
amistre64
  • amistre64
there was a maclaurin method ... taking a derivative and then integrating.
anonymous
  • anonymous
i start next week good luck here is what i did http://www.wolframalpha.com/input/?i=sum+%28-1%29^k*1%2Fk
anonymous
  • anonymous
now that we know it is \(-\ln(2)\) the first thing is to think of the expansion of \[\ln(1+x)\]
amistre64
  • amistre64
if we take the derivative assuming we started at x instead of 1 1 - x + x^2 - x^3 + ... looks functionally familiar
anonymous
  • anonymous
or what @amistre64 is doing, think of the anti derivative of \(\frac{1}{1+x}\)
amistre64
  • amistre64
gotta run :)
anonymous
  • anonymous
have fun!
anonymous
  • anonymous
I will
terenzreignz
  • terenzreignz
You seem to have attracted quite the OS users XD
terenzreignz
  • terenzreignz
Were you actually supposed to use a power series approach here? :D
anonymous
  • anonymous
I don't know, this wasn't for a specific class
anonymous
  • anonymous
I was just wondering what would happen if you used a summation and an improper integral together
anonymous
  • anonymous
I thought it would be interesting, and indeed it was.
terenzreignz
  • terenzreignz
Then you already knew how to do this? XD
anonymous
  • anonymous
not this part, I made up the problem of simple pieces and ended up with more than I bargained for
anonymous
  • anonymous
First I tried \[\sum_{k=1}^{\infty} \int\limits_{k}^{\infty} \frac{1}{x^2}dx\] then I multiplied the integral by (-1)^k when it didn't converge. I knew \[\sum_{k=1}^{\infty} (-1)^k \frac{ 1 }{ k }\] converged, I just didn't know how to figure out what it converged to.
terenzreignz
  • terenzreignz
Argh... not my day today.. \[\Large \frac1{1+t}= 1-t+t^2-t^3+t^4-t^5...\]
terenzreignz
  • terenzreignz
or\[\Large \frac1{1+t}=\sum_{k=0}^\infty(-1)^kt^k\]
terenzreignz
  • terenzreignz
On second thought, this doesn't really make sense since that right-series doesn't converge when t=1.
anonymous
  • anonymous
oh. the taylor series approximation for ln(1+x) looks a lot like the sum I found.
terenzreignz
  • terenzreignz
Checking...
terenzreignz
  • terenzreignz
Trying to derive...
anonymous
  • anonymous
http://en.wikipedia.org/wiki/Natural_logarithm#Derivative.2C_Taylor_series
terenzreignz
  • terenzreignz
doesn't show derivation though... I'm going to try and do it XD
anonymous
  • anonymous
http://en.wikipedia.org/wiki/Mercator_series#Derivation
terenzreignz
  • terenzreignz
Entirely different derivation... I'm trying Taylor...
terenzreignz
  • terenzreignz
Okay... done, I'm convinced XD

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