I don't get it

I don't get it

-1/12=-1/12

they both equal 2

[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]

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

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

your first line is appaling

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

excuse me wtf does sigma with infinite hat mean?

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

2=2

[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]

[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

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.

[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]

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

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

Are you a retard?

2

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

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

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

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?

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

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

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