an is a sequence of positive reals decreasing monotonically to zero. bn is defined by bn = an - 2an+1 + an+2 and all bn are non-negative. Prove that b1 + 2b2 + 3b3 + ... = a1.
Solution
We have that b1 + 2b2 + 3b3 + ... + n bn = a1 - (n+1)an+1 + nan+2 = a1 - (n+1)(an+1 - an+2) - an+2. Also we are given that all bi are non-negative. So S n bn is monotonic increasing and bounded above by a1. So it converges to a limit L ≤ a1.
bn = (an - an+1) - (an+1 - an+2). Hence bn + bn+1 + bn+2 + ... + bn+m = (an - an+1) - (an+m+1 - an+m+2). But am tends to zero, so (an+m+1 - an+m+2) tends to zero as m tends to infinity. Hence bn + bn+1 + bn+2 + ... converges to an - an+1. Hence (n+1)(bn+1 + bn+2 + bn+3 + ... ) converges to (n + 1)(an+1 - an+2). Hence L ≥ b1 + 2b2 + 3b3 + ... + n bn + (n+1)(bn+1 + bn+2 + bn+3 + ... ) = a1 - (n+1)an+1 + n an+2 + (n+1)(an+1 - an+2) = a1 - an+2. But an+2 tends to zero. Hence L = a1.
© John Scholes
jscholes@kalva.demon.co.uk
11 Mar 2002