Let a and b be positive real numbers, with a>b. Compute \(\frac{1}{ba} + \frac{1}{a(2a - b)} + \frac{1}{(2a - b)(3a - 2b)} + \frac{1}{(3a - 2b)(4a - 3b)} + \dotsb.\)

Guest Dec 8, 2018

#1**+1 **

Take a subsequence of N terms of this.

Split each term up using partial fractions

You'll see this reveals that the sequence is a telescoping sequence.

After the sum only part of the Nth term remains

\(\dfrac{N}{b (a N+b (-N)+b)}\)

Taking the limit of this as N goes to infinity results in

\(\dfrac{1}{b(a-b)}\)

.Rom Dec 8, 2018