Find the sum to n terms of the series `(1)/(1*3*5*7*9)+(1)/(3*5*7*9*11)+(1)/(5*7*9*11*13)+"......"`. Also, find the sum to infinty terms.

Explanation

Let `T_(n)` be the nth term of the given series.
then, `T_(n)=(1)/((2n-1)(2n+1)(2n+3)(2n+5)(2n+7)) " ".........(i)"`
`therefore V_(n)=(1)/((2n+1)(2n+3)(2n+5)(2n+7))`
[leaving first factor from denominator of `T_(n)` ]
` V_(n-1)=(1)/((2n-1)(2n+1)(2n+3)(2n+5))`
` implies V_(n)- V_(n-1)=(1)/((2n+1)(2n+3)(2n+5)(2n+7))-(1)/((2n-1)(2n+1)(2n+3)((2n+5))`
`=((2n-1)-(2n+7))/((2n-1)(2n+1)(2n+3)(2n+5)(2n+7))`
`=-8T_(n) " " " " [" from Eq. (i) "]`
`:. T_(n)=-(1)/(8)(V_(n)-V_(n-1))`
`:. S_(n)=sumT_(n)=sum_(n=1)^(n)T_(n)=-(1)/(8)sum_(n=1)^(n)(V_(n)-V_(n-1))=-(1)/(8)(V_(n)-V_(0))`
`[ " from important Theorem 1 of " sum]`
`=(1)/(8)(V_(0)-V_(n))`
`=(1)/(8){(1)/(1*3*5*7)-(1)/((2n+1)(2n+3)(2n+5)(2n+7))}`
`=(1)/(840)-(1)/(8(2n+1)(2n+3)(2n+5)(2n+7))`
and `S_(infty)(1)/(840)-(1)/(infty)=(1)/(840)-0=(1)/(840)`
Shortcut Method
`(1)/(1*3*5*7*9)+(1)/(3*5*7*9*11)+(1)/(5*7*9*11*13)+"......"+(1)/((2n-1)(2n+1)(2n+3)(2n+5)(2n+7))"......(i)"`
Now, in each term in denominator
`9-1=11-3=13-5="...."=(2n+7)-(2n-1)=8`
ThenEq. (i), can be written as
`=(1)/(8){(9-1)/(1*3*5*7*9)+(11-3)/(3*5*7*9*11)+(13-5)/(5*7*9*11*13)+"......"+((2n+7)-(2n-1))/((2n-1)(2n+1)(2n+3)(2n+5)(2n+7))}`
`=(1)/(8){(1)/(1*3*5*7)-(1)/(3*5*7*9)+(1)/(3*5*7*9)-(1)/(5*7*9*11)+(1)/(5*7*9*11)-(1)/(7*9*11*13)+"......"+(1)/((2n-1)(2n+1)(2n+3)(2n+5))-(1)/((2n+1)(2n+3)(2n+5)(2n+7))}`
`=(1)/(8){(1)/(1*3*5*7)-(1)/((2n+1)(2n+3)(2n+5)(2n+7))}`
[middle terms are cancelled out]
`=(1)/(840)-(1)/(8(2n+1)(2n+3)(2n+5)(2n+7))=S_(n) " " [" say "]`
`:.` Sum to infinity terms `=S_(infty)=(1)/(840)-0=(1)/(840)`.