I don't get it

>>8139682
-1/12=-1/12

>>8139684
they both equal 2

>>8139682
[math]\Sigma \frac{n}{2^n} = \Sigma n (\frac{x}{2})^n |_{x=1} = \frac{x}{2}\Sigma n (\frac{x}{2})^{n-1} |_{x=1} = x\frac{d}{dx}\Sigma (\frac{x}{2})^n |_{x=1} = x\frac{d}{dx}\frac{2}{2-x} |_{x=1} = x\frac{2}{(2-x)^2} |_{x=1} = 2 [/math]

[math]\Sigma \frac{2}{2^n} = \frac{2}{2-x}|_{x=1} = 2 [/math]

>>8139682
Intuitively, on the left hand side, the very initial numerator is 2, which is greater than the 1 on the right hand side, and this initial numerator on either side also has the smallest denominator, making it the greatest contributing term. So then even though all of the numerators thereafter are greater on the right hand side, the denominators increase rapidly enough to make this not surpass that initial gain that the left side has initially.

aw shit that's pretty cool

>>8139682
i don't get it, this breaks down for n = 1.

>>8139695
your first line is appaling

>>8139682
>doesn't know that n = 2
enjoy your mcjob

excuse me wtf does sigma with infinite hat mean?

>>8139765
The sum of the series to infinity. Did you take algebra 2?

>>8139682
2=2

>>8139682
[math]S = \sum_{1}^{\infty} \frac{n}{2^n} , \sum_{1}^{\infty} \frac{1}{2^n} = 1 ,
S + 1 = \sum_{1}^{\infty} \frac{n + 1}{2^n} = 2 \sum_{1}^{\infty} \frac{n + 1 }{2^{n + 1}} = 2 (\sum_{2}^{\infty} \frac{n}{2^n}) = 2 (S - \frac{1}{2}) = 2S - 1[/math]

File: Jessica_Clements .jpg (124KB, 750x723px) Image search: [Google]

124KB, 750x723px

[math] \int_{-a}^a f(x^2) \dfrac{ 1 } { 1 + {\mathrm e}^{ x^2 \sin(x) } } \, {\mathrm d} x = \int_0^a f(x^2) \, {\mathrm d}x [/math]

spooky

>>8139682

I guess if they converge to the same thing but a specific case like n=1 shows that the equality does not hold.

Applying simple convergence tests and you can see they both converge to 2. Geometric for the first infinite series and ratio for the second.

>>8139682

[eqn]\sum_{n=1}^\infty \frac{n}{2^n} = \sum_{n=1}^\infty n + \sum_{n=1}^\infty \frac{1}{2^n} = \frac{-1}{12} + 1 = \frac{11}{12}[/eqn]

>>8139770
got good grades until calc 5 years ago. About to relearn from prealgebra and go as far as i can

>>8139854
How is n/2^n = n + 1/2^n?

>>8139716
Are you a retard?

>>8139682
2<n (not true for all n in N)

>>8139682
[math]\sum_{n=1}^m2^{-n+1}-n*2^{-n} = 2^{-m} m[\math]

>>8140661
>[math]\sum_{n=1}^m2^{-n+1}-n*2^{-n} = 2^{-m} m[/math]

>>8139682
you'd troll better if it was from n=0 with a 1 instead of a 2

>>8139682
>>8139695
>>8139794
I don't lurk this board very much (but I do have a math degree). Is this just some kind of meme wherein mathfags (or perhaps even just texfags) try to fuck with non-mathfags? Did I just ruin the joke?

>>8140117
It's a joke. Most of this thread is.

>>8140676
Like anybody you can re-learn the converge tests and try it yourself.

>>8140734
It's not a meme it's just a thing where people fuck with eachother constantly