Question

Help! How do I approach this question:
Obtain an explicit formula for function(x) of
f(x)=sum when n=1 to infinity of (nx^n)

Answer 1

@Jhannybean

Answer 2

Recall that for \(|x|<1\), you have
\[\begin{align*}\sum_{n=0}^\infty x^n&=1+x+x^2+\cdots\\
&=\frac{1}{1-x}
\end{align*}\]
Suppose you take the derivative:
\[\begin{align*}\sum_{n=0}^\infty nx^{n-1}&=\frac{d}{dx}\left[\frac{1}{1-x}\right]\\\\
&=\frac{1}{(1-x)^2}
\end{align*}\]
Multiply both sides by \(x\):
\[\begin{align*}x\sum_{n=0}^\infty nx^{n-1}&=\sum_{n=0}^\infty nx^n\\\\
&=\sum_{n=1}^\infty nx^n
\end{align*}\]

Answer 3

small correction there
\[\frac{ d }{ dx }(\frac{ 1 }{ 1-x })=\frac{ 1 }{ (x-1)^2 }\]

Answer 4

@mathmath333 we have the same result :)
\[\frac{1}{(x-1)^2}=\frac{1}{[(-1)(1-x)]^2}=\frac{1}{(1-x)^2}\]

Answer 5

oh yes how can i miss that, lol

Answer 6

Thank you guys!

Answer 7

or

let \[S=x+2x^2+3x^3+\cdots\]

then \[xS=x^2+2x^3+3x^4+\cdots\]
subtract

\[S-xS=x+x^2+x^3+x^2+\cdots=\sum_{n=1}^{\infty}x^n=x\sum_{n=1}^{\infty}x^{n-1}=\frac{x}{1-x}\]
\[S(1-x)=\frac{x}{1-x}\]

\[S=\frac{x}{(1-x)^2}\]

Answer 8

cool

Answer 9

no one pointed out my typo

\[S-xS=x+x^2+x^3+x^{\color{red}{4}}+\cdots=\sum_{n=1}^{\infty}x^n=x\sum_{n=1}^{\infty}x^{n-1}=\frac{x}{1-x}\]