Suppose {f_n}{fn} and {g_n}{gn} 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))N∑n=0fn(gn+1−gn)=fN+1gN+1−f0g0−N∑n=0gn+1(fn+1−fn).
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))N∑n=0fn(gn+1−gn)+gn+1(fn+1−fn).
Expanding gives,
sum_(n=0)^N f_ng_(n+1)-f_ng_(n)+g_(n+1)f_(n+1)-g_(n+1)f_(n)N∑n=0fngn+1−fngn+gn+1fn+1−gn+1fn,
sum_(n=0)^N f_(n+1)g_(n+1)-f_ng_nN∑n=0fn+1gn+1−fngn.
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)N∑n=0fn+1gn+1−fngn=fN+1gN+1−f0g0.
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)N∑n=0fn(gn+1−gn)+gn+1(fn+1−fn)=fN+1gN+1−f0g0,
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))N∑n=0fn(gn+1−gn)=fN+1gN+1−f0g0−N∑n=0gn+1(fn+1−fn) as required.
Then, let f_(n)=nfn=n and let g_(n+1)-g_(n)=2^(-n)gn+1−gn=2−n. We see that if g_(n)=k2^(-n)gn=k2−n then
g_(n+1)-g_(n)=k(2^(-(n+1))-2^(-n))gn+1−gn=k(2−(n+1)−2−n),
g_(n+1)-g_(n)=k(1/2*2^(-n)-2^(-n))gn+1−gn=k(12⋅2−n−2−n),
g_(n+1)-g_(n)=-(k)/2 2^(-n)gn+1−gn=−k22−n.
Then let k=-2k=−2 giving g_(n+1)-g_(n)=2^(-n)gn+1−gn=2−n for g_(n)=-2*2^(-n)gn=−2⋅2−n
We conclude f_(n)=nfn=n, g(n)=-2*2^(-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)N∑n=0n2n=−2(N+1)2−(N+1)+2N∑n=02−(n+1)(n+1−n),
sum_(n=0)^N n/2^n = -2(N+1)2^(-(N+1))+sum_(n=0)^(N)2^(-n)N∑n=0n2n=−2(N+1)2−(N+1)+N∑n=02−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)N∑n=0n2n=−2(N+1)2−(N+1)+1−2−(N+1)1−12,
sum_(n=0)^N n/2^n = 2-2*2^(-(N+1))((N+1)+1)N∑n=0n2n=2−2⋅2−(N+1)((N+1)+1),
sum_(n=0)^N n/2^n = 2 - (N+2)2^(-N)N∑n=0n2n=2−(N+2)2−N
As N -> +\inftyN→+∞ (N+2)2^(-N) -> 0(N+2)2−N→0.
Then
sum_(n=0)^(\infty) n/2^n = 2∞∑n=0n2n=2.
