Reversing Taylor Series to find the original function

[delta59]

Homework Statement

I need to find the convergence a unknown function. Now I know the Taylor series of it which is 1/3+2/(3^2)+3/(3^3+4/(4^4+...+k/(3^k). Which mean I can just take the Riemann sum of k/(3^k) from say 0 to 50 and that would give me 3/4.

However this is not enough I need to confirm the answer by finding the unknown function. Normally this isn't too difficult as I would manipulate the series to get it into some sort of form that is familiar like e^x or cos(x), but this one is throwing me for a loop.

The bottom part (3^k) fits x/(1-x) when x=3 but I am not sure about the top

any help or hints would be great!

 

[lazyaditya]

I don't know if i am write or wrong , as i have never dealed with them but can't we write them as [summation of (K-1)/3^K]

 

[I like Serena]

Homework Statement

I need to find the convergence a unknown function. Now I know the Taylor series of it which is 1/3+2/(3^2)+3/(3^3+4/(4^4+...+k/(3^k). Which mean I can just take the Riemann sum of k/(3^k) from say 0 to 50 and that would give me 3/4.

However this is not enough I need to confirm the answer by finding the unknown function. Normally this isn't too difficult as I would manipulate the series to get it into some sort of form that is familiar like e^x or cos(x), but this one is throwing me for a loop.

The bottom part (3^k) fits x/(1-x) when x=3 but I am not sure about the top

any help or hints would be great!

Welcome to PF, delta59!

Perhaps you could start with substituting for instance x=1/3?
What would you get?

Afterward, try to think of defining a suitable function f(x) of which you know what the series sum is if you integrate f(x).
(I'll explain later.)

 

[delta59]

x/(1-x) when x=1/3 gives the correct bottom but then I just need to square the bottom function to get the top
so the f(x)=x/((1-x))^2 gives the expansion of 1/3+2/(3^2)+3/(3^3+4/(4^4+...+k/(3^k)

Thanks bouncing ideas I got now.

 

[I like Serena]

Yep. That works too! :)