Question

1 - 1/2 + 1/3 - 1/4 + 1/5 ...........upto infinity.

Answer 1

@dumbcow I need a method to do this. I have one. Help?

Answer 2

@TuringTest

Answer 3

@experimentX

Answer 4

alternating series, converges conditionally

Answer 5

How would you prove/show it?

Answer 6

do you want to calculate sum or prove it converges/ diverges ??

Answer 7

Calculate its value.

Answer 8

@SomeoneYouUsedToKnow : That is NOT the general term.
General term : (-1)^(r-1) / r

Answer 9

I have thought of one solution, need another perhaps better one.

Answer 10

1 - (1/2 + 1/4 + 1/8 + ...) + (1/3 + 1/9 + 1/27+...) +(1/5 + 1/25+ ...)- (1/6 + 1/36+ ..) + ...

hmm i don't think it's gonna work :/ lemme think of another one

Answer 11

there's a tricky method

Answer 12

http://home.scarlet.be/math/taylor.htm#ln(1+x)-and-its-Macl

Answer 13

Okay. Mine was the Geometrical series.
@experimentX Which is?

Answer 14

ln(1+x) = x/1 - x2/2 + x3/3 - x4/4 + ... + (-1)n-1.xn/n

now put x=1

Answer 15

Ohhhhhhh. I see.

Answer 16

Thats what I got.
Usint Infinite Geometrical series, Common ratio -x. Integrate from 0 to 1.

Answer 17

@experimentX how do you know that result directly?

Answer 18

usually do these stuffs these days.

Answer 19

i would like to see your method.

Answer 20

Okay.
1 - x + x^2 - x^3 ................infinity = 1 / 1+x
Integrateit from 0 to 1.
x - x^2 /2 +x^3 /3 - x^4 /4 ..........|0to1 = ln(1+x) |0to1
Putting the values We get req. sum inLHS and ln2 in RHS

Answer 21

Does that make sense?

Answer 22

ohhh nice sharan, nice

Answer 23

of course, it does.

Answer 24

are you preparing for the olympiads?

Answer 25

Haha. IIT. you?

Answer 26

another way of doing ...that's great!!!

Answer 27

Haha. Thanks. I thought maybe Just maybe you could do it without any calculus.

Answer 28

IIT. and IIIT.

Answer 29

Try this book if you guys like solving math problems. I haven't solved it yet, but I do like it.

http://blngcc.files.wordpress.com/2008/11/paulo-ney-de-souza-jorge-nuno-silva-berkeley-problems-in-mathematics.pdf

Answer 30

Ahaa. Thanks.
IIT and IIT Lol @ that. :D