Proving a series diverges 1/n and 1/n^2 = PI^2/6 .:Resolved:.

[voidflux]

Hello everyone!
I was wondering if anyone happend to know any links or just knew off the top of their head to show the following:
I need to show that this series diverges:
E shall stand for Sumnation

infinity
E          1/n  = 1+1/2+1/3+1/4+...+infinity
n = 1

How am i suppose to show that this series
 does infact diverge? I can't write, well its 
going to go to infinity, well acutally, i thought 
it would go to 0, but thats not the case.

Also some math wiz figured out that

infinity
E         1/n^2  = PI^2/6
n =1

anyone know how i could show this?

Thanks!

[alkatran]

If the increase on each level is greater than the decrease of the increase then it diverges (assuming that means it can reach infinity)

1 = 1

1 + 1/2 = 1.5
1.5-1 = .5, .5/1 = .5

1 + 1/2 + 1/3 = 1.833333
1.8333-1.5 = .33333, .3333/.5 = .66666

So, the decrease in the increase is actually going down each level.. so it will reach forever (if you did 1 +1/2 + 1/4 the decrease is constant)

Honestly, I'm making this all up as I go, so I might not be right.

[alkatran]

oh, and for the pi one... area of circle = pi*r^2.. I'm guessing that has something to do with it. Now.. to play with the formula (weee)

b=0
for a = 1 to infinity
b=b+1/a^2
next a

b = pi^2/6

so...

b=0
for a = 1 to infinity
b=b+6/a^2
next a

b = pi^2


I can see it in my mind... it gets bigger then smaller... but I don't know how to prove it! I mean, how do you break down a seri- whhhoooooahhh

nm... it didn't work because we're dividing.
ok.. Usually when trying to sum a series you figure out the average then multiply that by the amount... yep.. out average is right around 0 and our amount is infinity..

the series definitly stops increasing before it crosses pi^2 but I dunno...

back to area idea

AreaC = Pi*r^2
AreaC^2/r^4 = Pi^2

AreaC^2/r^4 = sumofthatseries


argh.. I give up

[Something Else]

Hmm, lets see:

1 + 1/2 + 1/3 + 1/4 + ...=>
can be said to be made up at least partially by:

1/2 + 1/4 + 1/8 + ...
1/6 + 1/12 + 1/24 + ...
1/10 + 1/20 + 1/40 + ...

or

(1/2 + 1/4 + 1/8 + 1/16)*(1 + 1/3 + 1/5 + ... + 1/p)

now 1/2 + 1/4 + 1/8 + ... = 1
so

1/2 + 1/4 + 1/8 + ...
1/6 + 1/12 + 1/24 + ...
1/10 + 1/20 + 1/40 + ...
= (1 + 1/3 + 1/5 + ... + 1/p)

so 1 + 1/2 + 1/3 + 1/4 + ... can be said to be:

1 + (1 + 1/3 + 1/5 + ... + 1/p) + what hasn't been considered, or:

everything thats not 1/2^n, or (1/p)*1/2^n

Well, theres (1/p)^n, oops, gota go.

[voidflux]

Thanks guys, feel free to continue to write your idea's!

[voidflux]

i discovered the solution, for the first one you can just use the integral test,
let f(x) = x^-1

integral from 1 to b as x->infinity, integrate it, and u'll get
ln|b|-ln|1| = infinity

for the 2nd one, u can write:

k 1+1/4+....+1/k^2
5 1.46361
10 1.54977
10^2 1.63498
10^3 1.64393
10^4 1.64483
10^5 1.64492
10^6 1.64493

and
PI^2
-----
6

is equal to 1.6449

thanks for the help everyone