infinite sum of 3/e^n + 2/(n(n+1))

Question

I evaluated the first part as a geometric series and the second part with an integral test.

I got r = 1/e for the first part and -2ln(2) for the integral of the second part. I'm not sure how to find the sum of this convergent series

Answer 1

2/[n(n + 1)] = A/n + B/(n + 1)  (partial fraction decomposition)

 

A(n + 1) + Bn = 2

 

(A + B)n + A = 0n + 2    A + B = 0   and  A = 2

 

A = 2 and B = -2

 

So, 2/[n(n + 1)] =  2/n - 2/(n+1)

 

The sum of the first n terms of the series with nth term 2/[n(n+1)] is S_n = 

 

(2-1) + (1-2/3) + (2/3-2/4) + (2/4-2/5) + ... + (2/n - 2/(n+1)]

 

= 2 - 2/(n + 1)   (everything cancels except for the first and last)

 

Sum of series = limit as n→∞ {S_n}  = 2

 

The first series is geometric with r = 1/e.  Since -1 < r < 1, the series has sum a/(1-r), where a is the first term of the series.

 

a/(1-r) = (3/e) ÷ (1-1/e) = (3/e)[e/(e-1)] = 3/(e-1)

 

∑ from 1 to ∞ {3/eⁿ + 2/[n(n+1)]} = 3/(e-1) + 2

 

                                                                = (2e + 1)/(e - 1)