Is the series convergent or divergent?

Question

Determine whether the series is convergent or divergent by expressing Sn as a telescoping sum.
#sum_(n=5)^oo6/(n^2-1)#

If it is convergent, find its sum.

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Answer 1 · 2017-10-07T01:51:49

#27/20#

#sum_(n=5)^oo6/(n^2-1)#

First, express #6/(n^2-1)# using partial fractions.
#6/(n^2-1)=A/(n-1)+B/(n+1)#
#6=A(n+1)+B(n-1)#

Substitute #n=1# to get #A=3# and #n=-1# to get #B=-3#.

Thus, #sum_(n=5)^oo6/(n^2-1)=sum_(n=5)^oo3/(n-1)-3/(n+1)#.

Rewrite this using limit notation: #lim_(L->oo)sum_(n=5)^L3/(n-1)-3/(n+1)#.

Find the first few terms: #lim_(L->oo)((3/4-3/6)+(3/5-3/7)+(3/6-3/8)+(3/7-3/9)+\ldots+(3/(L-1)-3/(L+1)))#.

Rearrange the terms: #lim_(L->oo)(3/4+3/5+(-3/6+3/6)+(-3/7+3/7)+(-3/8+3/8)+\ldots+(-3/(L-1)+3/(L-1))+(-3/L+3/L)-3/(L+1)))#.

Most of the terms cancel out, leaving #lim_(L->oo)3/4+3/5-3/(L+1)#.

This equals #27/20#.