How do you find the nth partial sum, determine whether the series converges and find the sum when it exists given #1/(13)+1/(35)+1/(5*7)+...+1/((2n-1)(2n+1))+...#?

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Answer 1 · 2017-03-19T06:19:42

The nth partial sunm is #=n/(2n+1)#
The series converge to #1/2#

Let's perform the partial fraction decomposition of the nth term

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

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

The numerators are the same, we compare the denominators

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

Let #n=1/2#, #=>#, #1=2A#, #=>#, #A=1/2#

Let #n=-1/2#, #=>#, #1=-2B#, #=>#, #B=-1/2#

Therefore,

#1/((2n-1)(2n+1))=(1/2)/(2n-1)-(1/2)/(2n+1)#

#u_n=1/2(1/(2n-1)-1/(2n+1))#

So,

#u_1=1/2(1/1-cancel(1/3))#

#u_2=1/2(cancel(1/3)-cancel(1/5))#

#u_3=1/2(cancel(1/5)-cancel(1/7))#

#u_(n-1)=1/2(cancel(1/(2n-3))-cancel(1/(2n-1)))#

#u_n=1/2(cancel(1/(2n-1))-1/(2n+1))#

#sum_(n=1) ^n=1/2(1-1/(2n+1))=1/2((2n)/(2n+1))#

#=n/(2n+1)#

The nth partial sunm is #=n/(2n+1)#

#lim_(n->+oo)n/(2n+1)=lim_(n->+oo)n/(2n)=1/2#

The series converge to #1/2#

How do you find the nth partial sum, determine whether the series converges and find the sum when it exists given #1/(1*3)+1/(3*5)+1/(5*7)+...+1/((2n-1)(2n+1))+...#?