Compute closed form of $\sum_{k = 1}^{\infty} \frac{k}{3^{k}}$ Compute closed form for: $$f(x)  = \sum_{k = 1}^{\infty} \frac{k}{3^{k}}$$
EDIT: There appears to have been solutions linked to this question as it is seen a s duplicate. One of them is a bit advanced and doesn't really explain the intermediate steps that some of the less mathematically mature may have difficulties piecing together. I feel this one may benefit beginners in Real Analysis more.
This question is in a chapter on power series and uniform convergence.
My Ideas:
When I look at this what comes to mind is that in order to find a closed form solution I'm going to have to attempt to manipulate this expression into a power series from that tools are available to use to compute it.
That being the case I would like to get this into something like: $$\sum_{k = 1}^{\infty}a_{k}x^{k}$$.
I see if I take the integral of both sides I would arrive at something of the form:
$$\int_{0}^{x}f(t) dt  = \int_{0}^{x} \sum_{k = 1}^{\infty} \frac{k}{3^{k}}dt \\ \Rightarrow \ = \sum_{k = 1}^{\infty} \frac{k}{3^{k}}x$$
If I take this integral $k$ times I would be able to get a $x^k$ term eventually. But I don't see anything fruitful coming from that, maybe I would have to play around with it more.
Another thought was examining the $\frac{k}{3^k}$ term I ask myself if there is a time that this ratio would ever be greater than $1$, because if not then I may have my geometric series right in front of me. Attempting to figure when $\frac{k}{3^{k}} > 1$, the only thing I arrived at was: $$\frac{\ln{k}}{k} > \ln{3}$$.
Which isn't very conclusive. So assuming that it is always true that  $\frac{k}{3^{k}} > 1$, the closed from of the solution would be:
$$\frac{3^k}{3^k - k}$$
Feedback on the thought process? Am I heading in the right direction with either idea? or perhaps something else should be tried? 
 A: We know that if $|x|<1$ then $\sum_{k=0}^\infty x^k=\frac{1}{1-x}$. We can differentiate this power series term by term in $(-1,1)$. So by differentiating we get $\sum_{k=1}^\infty kx^{k-1}=\frac{1}{(1-x)^2}$. If we multiply both sides by $x$ we get $\sum_{k=1}^\infty kx^k=\frac{x}{(1-x)^2}$. Now put $x=\frac{1}{3}$. 
A: Hint: let 
$$
g(x) = \sum_k k x^k.
$$
What is $g(\frac{1}{3})$? Can you find a general expression for $g(x)$? What about for $h(x) = x g(x)$...that might be simpler, esp. if you looked at an integral. 
A: I don't know about how to solve this by integrals and all but there is a concept known and arithmetico-geometric series.
This problem is where the series goes on till infinity.
General form of arithmetico-geometric series is 
a.r +(a+d).r^2 +(a+2d).r^3 + (a+3d).r^4 +......
Basically to solve this you multiply the sum by the common ratio. 
After that you can rearrange it and subtract it by the sum itself so that you can take the common diff. Common out of it(I mean distributive property).
Then by using sum of infinite gp, it can be solved.
In this problem
S    = 1/(3^1)+(1+1)/(3^2)+(1+2.1)/3^3+....
S/3=.                1/3^2.        +  (1+1)/3^3
2S/3= 1/3+ 1/3^2+1/3^3+...
2S/3= (1÷3)/(1-1/3)
2S/3= (1÷3)/(2÷3)
2S/3= 0.5
S = 0.75
Pardon for really bad typing and correct me if I am wrong
