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))+...1/(1*3)+1/(3*5)+1/(5*7)+...+1/((2n-1)(2n+1))+...?