Let a and b be positive real numbers, with a>b. Compute \(\frac{1}{ba} + \frac{1}{a(2a - b)} + \frac{1}{(2a - b)(3a - 2b)} + \frac{1}{(3a - 2b)(4a - 3b)} + \dotsb.\)
+6248 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
\(\dfrac{1}{b(a-b)}\)
