MP

[Jetru]

Maths problem.
1/3 +1/(3^2).2 +1/(3^3).3+....

Sum it. General term is 1/(3^x).x
.=*

I tried to integrate...Couldn't pull it off. Is that correct anyway?

edit--
Alright, tried this is Mathematica,it said:
ExpIntegralEi[-x Log[3]]

The function reference said Ei(z)=-Int_-x to inf. _(e^-t)/t dt
I don't quite get it.

Another ques:In Mathematica, the 'd' in 'dx' looks like 'dl'. Why? Is that the actual way of writing the small change 'd'?

[Alvaro]

General term is 1/(3^x).x
.=*

Do you mean (1/(3^x))*x or 1/((3^x)*x)?

[schloob]

General term is 1/(3^x).x
.=*

Do you mean (1/(3^x))*x or 1/((3^x)*x)?

my guess is 1/((3^x)*x) because otherwise he probably would've put like x/3^x

[Alvaro]

log(3/2)

[Jetru]

Why don't you ever tell how?

[Alvaro]

Plug in x=-1/3 in the Taylor expansion of log(1+x) and you'll get that your number is -log(2/3), which is also log(3/2).

[Jetru]

Don't know Taylor series. Is there a way to do it without knowing the taylor series?

[Jetru]

Okay, i got it. This Taylor thing seems weird.

[Alvaro]

Okay, i got it. This Taylor thing seems weird.

Well, there was a professor in my university that used to say that 90% of Math is Taylor's theorem.

[DannyBoy]

Okay, i got it. This Taylor thing seems weird.

Well, there was a professor in my university that used to say that 90% of Math is Taylor's theorem.

I believe your professor, as it seems that most of the maths I do requires a Taylor series expansion at one point or another.