I need help on this problem. Show that the series 2/(n^2-1) from n=2 to infinity is convergent, and find its sum.

Question

I get stuck when using telescoping series. I appreciate whoever can help me. Thank you.

Answer 1

Use partial fractions:  2/(n²-1) = 2/{(n+1)(n-1)}

 

                                           = A/(n+1) + B/(n-1)

 

                                        2 = A(n-1)+B(n+1)

 

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

 

                                      So, A+B = 0

                                          -A+B = 2

 

                                              2B = 2   So, B = 1, A = -1

 

So, Σ_{(n=2 to ∞)}[2/(n²-1] = ∑_{(n=2 to ∞)}[-1/(n+1) +1/(n-1)]

 

Sum of first n terms =

 

S_n = (-1/3+1)+(-1/4+1/2)+(-1/5+1/3)+(-1/6+1/4) +...+                                                                                             (-1/(n+2)+1/n) 

 

    = 1 + 1/2 + 1/(n+1) - 1/(n+2)

 

lim_n→∞S_n = 3/2

 

The series converges to 3/2.  

 

Answer 2

The even terms are the following series

 

∑_k=2 1/[(2k)(k+1)],   where the sum goes k -> infinity      (**)

 

This telescopes as follows

 

1/[(2k)(k+1)] + 1/[2(k+1)(k+2)] = 2/[(2k)(k+2)], etc. giving

 

(**) = lim_j->∞ (1/4)(j/(2+j)) 

       = 1/4.

 

The odd terms telescope similarly, as you can show. They are written

 

∑_k=1 2/[(2k+1)(2k+3)],   where the sum goes to infinity

 

I hope that helps