Prove that the series x/(1+x)+2x^2/(1+x^2)+4x^4/(1+x^4)+8x^8/(1+x^8)+...?

Sum to infinity of a convergent geometric series is

a/(1 - r), where a is the first term, r is the common ratio and |r| < 1

So we have

x - x^2 + x^3 - x^4 + x^5 - x^6 + ... = x/(1 + x) .............. {1}

2x^2 - 2x^4 + 2x^6 - 2x^8 + 2x^10 - 2x^12 + ... = 2x^2 /(1 + x^2) .............. {2}

4x^4 - 4x^8 + 4x^12 - 4x^16 + 4x^20 - 4x^24 + ... = 4x^4 /(1 + x^4) .............. {3}

8x^8 - 8x^16 + 8x^24 - 8x^32 + 8x^40 - 8x^48 + ... = 8x^8 /(1 + x^8) .............. {4}

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Add all these equations up and simplify gives

x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8 + ...

Which is also in the form of a geometric series, hence its sum is

x/(1 - x)

Since {1}, {2}, {3}, {4}, and so on have common ratio of -x, -x^2, -x^4, -x^8 (and so on) respectively, and they converge when |x| < 1

Hence we have proved that it converges for |x| < 1 and its sum is x/(1 - x)

Another method from the point of view of generating functions:

\sum (2^n x^(2^n))/(1 + x^(2^n))

... = x \sum (2^n x^(2^n - 1))/(1 + x^(2^n))

... = x \sum D(log(1 + x^(2^n)))

... = x D(log(\prod (1 + x^(2^n))))

where D stands for differentiation with respect to x. Use induction on k to see that

(1 + x)(1 + x^2)(1 + x^(2^2))...(1 + x^(2^k)) = 1 + x + x^2 + x^3 + ... + x^(2^(k + 1) - 1)

As k (or n) tends to infinity, we therefore have

\prod (1 + x^(2^n)) = 1 + x + x^2 + x^3 + .... = 1/(1 - x)

for |x| < 1. Thus, for |x| < 1, it is composable with log and differentiable, and so the original series evaluates to

x D(log(\prod (1 + x^(2^n))))

... = x D(log(1/(1 - x)))

... = x(1 - x)/(1 - x)^2

... = x/(1 - x)

2 sheep

3 goats and

a jar of pickled onions

is my answer

Well i am not good at infinite series ,raabe test etc...

u just gave me a head ache