well?

>>9130156
Odd is given by:
N + N+2 + N+4...N+Infinity
Even is given by:
N+1 + N+3 + N+5...N+Infinity
furthermore if you go into negative numbers then they also cancel out the 1 difference at any point,
Thus unless it specifices at a specific Nth term, they are equal

>>9130159
>if you go into negative
>unless it specifices at a specific Nth term
fucking retard can't even read

>>9130165
T. Spastic
That guy clearly answered the terms of the question retard
Fuck off to whence you came

>>9130168
>That guy
not fooling anyone samefag

For all partial sums, even is larger.

Sums of both diverge.

>>9130156
>well?
...well Lrn2bigger, fgt pls

>>9130156
Let A be the sum of odds and B the sum of evens.
Then A+B=-1/12 and A-B=1/4.
Solving gives A=1/6 and B=-1/3.
See >>9130177 for a more serious answer.

>>9130156
For the first series notice that
x = 1 + 3 + 5 + ...
2x = 2 + 6 + 10 + ...
4x = 4 + 12 + 20 + ...
8x = 8 + 24 + 40 + ...
and so on. Therefore
x + 2x + 4x + 8x + ... = 1 + 2 + 3 + ... = -1/12
but at the same time
x + 2x + 4x + 8x + ... = x*(1 + 2 + 4 + ...) = x*(-1)
so together we get x = 1/12.

Now the other series is simply
2 + 4 + 6 + ... = 2*(1 + 2 + 3 + ...) = -1/6


Since 1/12 > -1/6 we get
1+3+5+... > 2+4+6+...

>>9130156
n^2 < n^2+n

File: ddddddddddddddas.png (8KB, 1280x902px) Image search: [Google]

8KB, 1280x902px

>>9130199
You are literally this retared

>>9130193
I thought that
[eqn] \frac{1}{6} + \left( - \frac{1}{3} \right) = -\frac{1}{6} \\
\frac{1}{6} - \left( - \frac{1}{3} \right) = \frac{1}{2} [/eqn]

>>9130220
>can't use tex
>calls others retarded

>>9130232
do half an infinity sign in tex then retard

>>9130242
[math]\propto[/math]

>>9130275
>half an hour to do a 3/4 instead of 1/2
lol

>>9130193
solving actually gives A=\frac{1}{12} and B=-\frac{1}{6}

>>9130275
isn't that just alpha?

[math]\alpha[/math]

>>9130242
x

The other half is left as an exercise to the reader.

>>9130156
first one is odder

the typed out "one" is bigger than the symbol "1"

>>9130156
[math]\sum_{i \in \mathbb{N}} 2i = 2 \sum_{i \in \mathbb{N}} i > 2 \sum_{i \in \mathbb{N}} 2i = 4 \sum_{i \in \mathbb{N}} i > 4 \sum_{i \in \mathbb{N}} 2i = \dots
[/math]

>>9130321
>>9130199
>>9130193
Can somebody explain why things like 'if every nth element of sum A is greater than or equal to nth element of sum B then sum A is greater than or equal to sum B' are void when sums are infinite?
Is it just because it werks?

>>9132532
>when sums are infinite
Sums are always finite.
If you connect infinitely many terms with additions when you get a series and not a sum. Different rules apply to series than to sums.

>>9132547
okay, I mixed up terms indeed, my stupid.

So it's untrue that if series A of 1's goes to infinity, then series B of numbers that all are bigger than 1 should also go to infinity, but 'faster'?

>>9132560
[eqn] \sum_{k=1}^\infty 1 = - \frac{1}{2} [/eqn]
Nothing goes to infinity here.

>>9132596
why is it so?
Are series that much different from sums?

>>9130165
>>9130172
retard

>>9132604
he's trolling you