Q. `3/1^2` + `5/(1^2+2^3)` + `7/(1^2+2^3+3^3)` +....................`oo` is

A) 3

B) 4

C) 5

D) 6

### 2 Answers | Add Yours

`3/1^3+5/(1^3+2^3)+7/(1^3+2^3+3^3)+..........oo`

THe general term of the above series

`T_n=(2n+1)/(1^3+2^3+3^3+...+^3)`

`=(2n+1)/((n^2(n+1)^2)/4)`

`=(4(2n+1))/(n^2(n+1)^2)`

`=4{1/n^2-1/(n+1)^2}`

`T_1=4{1/1-1/2^2}`

`T_2=4{1/2^2-1/3^2}`

`................`

`T_n=4{1/n^2-1/(n+1)^2}`

`sum_(r=1)^nT_r=4(1-1/(n+1)^2)=(4(n^2+2n))/(n+1)^2`

`n->oo`

`sum_(r=1)^ooT_r=lim_(n->oo)(4n^2(1+2/n))/(n^2(1+1/n)^2)`

`=4`

**Thus correct answer is B**

Find the infinite sum `3/1+5/9+7/36+9/100+...`

The terms can be written as `(2n+1)/((n(n+1))/2)^2` or `(4(2n+1))/(n^2(n+1)^2)` .

The partial sum is `sum_1^m (4(2m+1))/(n^2(n+1)^2)=(4(m^2+2m))/((m+1)^2)`

The infinite sum is the limit of the partial sum as m goes to infinity:

`lim_(m->oo)(4(m^2+2m))/((m+1)^2)=4lim_(m->oo)(m^2+2m)/(m^2+2m+1)`

`=4lim_(m->oo)(1+2/m)/(1+2/m+1/m^2)`

`=4(1)=4`

----------------------------------------------------------------

The sum of the infinite series is 4

---------------------------------------------------------------

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes