Question

summation to infinity

Answer 1

\[\sum_{k=1}^{\infty} (-1)^k \frac{ 1 }{ k }\]

Answer 2

I don't know how to approach this. How would I start?

Answer 3

I could list out a bunch of terms and add them together in my calculator, but I don't like doing it that way.

Answer 4

it's the easy way, not the best way

Answer 5

try to split the positive and negatives ... is an idea

Answer 6

you trying to find the actual sum?

Answer 7

yes

Answer 8

jeez
i would first of all cheat
then look for a power series representation of a function

Answer 9

\[1-\frac12x+\frac13x^2-\frac14x^3\pm...~:~x=1\]

Answer 10

maybe start with an x?

Answer 11

yeah exactly but first i cheated so i would know where i was headed

Answer 12

ive got classes today .... so i wont be back for about 2 hours. My advice is to post sooner :)

Answer 13

there was a maclaurin method ... taking a derivative and then integrating.

Answer 14

i start next week
good luck
here is what i did
http://www.wolframalpha.com/input/?i=sum+%28-1%29^k*1%2Fk

Answer 15

now that we know it is \(-\ln(2)\) the first thing is to think of the expansion of
\[\ln(1+x)\]

Answer 16

if we take the derivative assuming we started at x instead of 1

1 - x + x^2 - x^3 + ... looks functionally familiar

Answer 17

or what @amistre64 is doing, think of the anti derivative of \(\frac{1}{1+x}\)

Answer 18

gotta run :)

Answer 19

have fun!

Answer 20

I will

Answer 21

<whistles...>
You seem to have attracted quite the OS users XD

Answer 22

Were you actually supposed to use a power series approach here? :D

Answer 23

I don't know, this wasn't for a specific class

Answer 24

I was just wondering what would happen if you used a summation and an improper integral together

Answer 25

I thought it would be interesting, and indeed it was.

Answer 26

Then you already knew how to do this? XD

Answer 27

not this part, I made up the problem of simple pieces and ended up with more than I bargained for

Answer 28

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.

Answer 29

Argh... not my day today..
\[\Large \frac1{1+t}= 1-t+t^2-t^3+t^4-t^5...\]

Answer 30

or\[\Large \frac1{1+t}=\sum_{k=0}^\infty(-1)^kt^k\]

Answer 31

On second thought, this doesn't really make sense since that right-series doesn't converge when t=1.

Answer 32

oh. the taylor series approximation for ln(1+x) looks a lot like the sum I found.

Answer 33

Checking...

Answer 34

Trying to derive...

Answer 35

http://en.wikipedia.org/wiki/Natural_logarithm#Derivative.2C_Taylor_series

Answer 36

doesn't show derivation though... I'm going to try and do it XD

Answer 37

http://en.wikipedia.org/wiki/Mercator_series#Derivation

Answer 38

Entirely different derivation... I'm trying Taylor...

Answer 39

Okay... done, I'm convinced XD