Show that sum x/2^x = 2 summation running 0 to infinity ?

2 Answers
Jul 25, 2017

sum_(n=0)^(\infty) n/2^n = 2.

Explanation:

Suppose {f_n} and {g_n} are two sequences. Then,

sum_(n=0)^N f_n (g_(n+1)-g_(n)) = f_(N+1)g_(N+1)-f_0g_0-sum_(n=0)^(N)g_(n+1)(f_(n+1)-f_(n)).

This can be proved by investigating
sum_(n=0)^N f_n(g_(n+1)-g_(n))+g_(n+1)(f_(n+1)-f_(n)).

Expanding gives,

sum_(n=0)^N f_ng_(n+1)-f_ng_(n)+g_(n+1)f_(n+1)-g_(n+1)f_(n),
sum_(n=0)^N f_(n+1)g_(n+1)-f_ng_n.

This is now a telescoping where all the terms cancel apart from the first and the last, giving

sum_(n=0)^N f_(n+1)g_(n+1)-f_(n)g_(n)=f_(N+1)g_(N+1)-f_(0)g_(0).

Substituting,

sum_(n=0)^N f_n(g_(n+1)-g_(n))+g_(n+1)(f_(n+1)-f_(n)) = f_(N+1)g_(N+1)-f_(0)g_(0),
sum_(n=0)^N f_n (g_(n+1)-g_(n)) = f_(N+1)g_(N+1)-f_0g_0-sum_(n=0)^(N)g_(n+1)(f_(n+1)-f_(n)) as required.

Then, let f_(n)=n and let g_(n+1)-g_(n)=2^(-n). We see that if g_(n)=k2^(-n) then

g_(n+1)-g_(n)=k(2^(-(n+1))-2^(-n)),
g_(n+1)-g_(n)=k(1/2*2^(-n)-2^(-n)),
g_(n+1)-g_(n)=-(k)/2 2^(-n).

Then let k=-2 giving g_(n+1)-g_(n)=2^(-n) for g_(n)=-2*2^(-n)

We conclude f_(n)=n, g(n)=-2*2^(-n).

Substituting,

sum_(n=0)^N n/2^n = -2(N+1)2^(-(N+1))+2sum_(n=0)^(N)2^(-(n+1))(n+1-n),
sum_(n=0)^N n/2^n = -2(N+1)2^(-(N+1))+sum_(n=0)^(N)2^(-n),

By the sum of a geometric series,

sum_(n=0)^N n/2^n = -2(N+1)2^(-(N+1))+(1-2^(-(N+1)))/(1-1/2),
sum_(n=0)^N n/2^n = 2-2*2^(-(N+1))((N+1)+1),
sum_(n=0)^N n/2^n = 2 - (N+2)2^(-N)

As N -> +\infty (N+2)2^(-N) -> 0.

Then

sum_(n=0)^(\infty) n/2^n = 2.

Jul 25, 2017

See below.

Explanation:

Defining S_n = sum_(k=0)^n x^k or

S_n = (x^(n+1)-1)/(x-1)

for abs x < 1 we have

S_oo = lim_(n->oo)S_n = -1/(x-1)

now

sum_(k=0)^n k x^k = x ((dS_n)/(dx)) and

sum_(k=0)^oo kx^k = x ((dS_oo)/(dx)) = x(d/dx)(1/(1-x)) = x/(x-1)^2

and now making x = 1/2 we obtain

sum_(k=0)^oo kx^k = 2