How to prove this equality  $ t(1-t)^{-1}=\sum_{k\geq0} 2^k t^{2^k}(1+t^{2^k})^{-1}$? 
Prove the equality $\quad t(1-t)^{-1}=\sum_{k\geq0} 2^k t^{2^k}(1+t^{2^k})^{-1}$.

I have just tried to use the Taylor's expansion of the left to prove it.But I failed.
I don't know how the $k$ and $2^k$ in the right occur. And this homework appears after some place with the $Jacobi$ $Identity$ in the book $Advanced$ $Combinatorics$(Page 118,EX10 (2)).
Any hints about the proof ?Thank you in advance.
 A: It is clearer  without the summation symbol. We want to find
$$\frac{t}{1+t}+\frac{2t^2}{1+t^2}+\frac{4t^4}{1+t^4}+\frac{8t^8}{1+t^8}+\frac{16t^{16}}{1+t^{16}}+\cdots.\tag{$\ast$}$$
Add $\dfrac{-t}{1-t}$ on the left. Note that 
$$\frac{-t}{1-t}+\frac{t}{1+t}=\frac{-2t^2}{1-t^2}.$$
But 
$$\frac{-2t^2}{1-t^2}+\frac{2t^2}{1+t^2}=\frac{-4t^4}{1+t^4}\quad\text{and}\quad  \frac{-4t^4}{1-t^4}+\frac{4t^4}{1+t^4}=\frac{-8t^8}{1+t^8}.$$ 
Continue. So adding $\dfrac{-t}{1-t}$ kills the sum $(\ast)$, and therefore our sum must be $\dfrac{t}{1-t}$. 
To put it another way, the sum $(\ast)$ is almost a telescoping series. All it needed was a little nudge. 
The calculation above is a formal manipulation. However, the series $(\ast)$ converges whenever $|t|&lt1$, so the formal manipulation gives the correct answer. 
A: This is all for $|t|&lt1$.  Start with the geometric series
$$\frac{t}{1-t} = \sum_{n=1}^\infty t^n$$
On the right side, each term expands as a geometric series 
$$\frac{2^k t^{2^k}}{1+t^{2^k}} = \sum_{j=1}^\infty 2^k (-1)^{j-1} t^{j 2^k}$$
If we add this up over all nonnegative integers $k$, for each integer $n$ you get
a term in $t^n$ whenever $n$ is divisible by $2^k$, with coefficient $+2^k$ when
$n/2^k$ is odd and $-2^k$ when $n/2^k$ is even.  So if $2^p$ is the largest power of $2$ that divides $n$, the coefficient of $t^n$ will be $2^p -\sum_{k=0}^{p-1} 2^k = 1$. 
A: Hint $\ $ Let $\rm\:N\to\infty\ $ in  $\rm\displaystyle\ \sum_{K\!\:=\!\:0}^{N-1}\!\ \frac{2^{\:\!K}\:\! t^{2^{\:\!K}}}{t^{2^K}\!+1} + \frac{t}{t-1}\: =\ \frac{2^{\:\!N}\:\! t^{2^{\:\!N}}}{t^{2^{\:\!N}}\!-1}\ =\: c\ t^{2^{\:\!N}} + \:\cdots\ $ by  telescopy.
Since $\rm\: t^{2^{\:\!N}} \to 0\:$ as $\rm\:N\to \infty,\:$ the desired formal power series equality follows.
See here for more on convergence of formal power series (beware many make errors here).
Remark $\ $ The telescopic proof is simply $\rm 2^{\:\!N}$ times below, for $\rm\:x = t^{2^N}$
$$\rm \frac{2\:x^2}{x^2-1} - \frac{x}{x-1}\: =\: \frac{x}{x+1}$$
A: You can use:
$$\frac{t}{t-1} = \sum_{n=1}^\infty{t^n}=t+t^2+(t^3+t^4)+(t^5 + ... + t^8)$$
And then use:
$$\sum_{n=2^k}^{n=2^{k+1}}{t^{n}} = \frac{t^{2^k}(t^{2^{k+1}}-1)}{t-1}$$
