# Find the sum of the infinite series (n goes from 1 to infinity) of [7/(3)^n) + (6/((n+3)(n+4))]I get that the series can be split into two because of the addition. From this I can figure out the...

Find the sum of the infinite series (n goes from 1 to infinity) of [7/(3)^n) + (6/((n+3)(n+4))]

I get that the series can be split into two because of the addition. From this I can figure out the sum of the first infinite series but not of the second.

### 1 Answer | Add Yours

You should separate the terms of the sum such that:

`sum_(n=1)^oo (7/(3^n) + 6/((n+3)(n+4))) = sum_(n=1)^oo (7/(3^n) ) + sum_(n=1)^oo (6/((n+3)(n+4)))`

Notice that the sum of the terms `7/3 + 7/3^2 + ... + 7/3^n` represents the sum of terms of geometric series, whose common ratio is `(1/3).`

`sum_(n=1)^oo (7/(3^n) ) = 7*(1 - (1/3)^n)/(1 - 1/3) = 21/2`

You should use partial fraction decomposition for the term `6/((n+3)(n+4))` such that:

`6/((n+3)(n+4)) = A/(n+3) + B/(n+4)`

`6 = An + 4A + Bn + 3B`

`6 = n(A+B) + 4A+3B`

`{(A+B = 0),(4A+3B=6):}=>{(A=-B),(-4B+3B=6):}=>B=-6, A=6`

`6/((n+3)(n+4)) = 6/(n+3)- 6/(n+4)`

`1/((n+3)(n+4)) = 1/(n+3) - 1/(n+4)`

Notice that giving values for n yields:

`n=1 => 1/4 - 1/5`

`n = 2 => 1/5 - 1/6`

..........

`n = n-1 => 1/(n+2) - 1/(n+3) `

`n = n => 1/(n+3) - 1/(n+4)`

..............

`n -> oo => lim_(n->oo) 1/(n+3) - 1/(n+4) = 0`

Hence, `sum_(n=1)^oo (1/(n+3) - 1/(n+4)) = 1/4 - 1/5 + 1/5 - 1/6 + ... + 1/(n+2) - 1/(n+3) + 1/(n+3) - 1/(n+4) + ....+0`

`sum_(n=1)^oo (1/(n+3) - 1/(n+4)) = 1/4`

**Hence, evaluating the given series, yields `sum_(n=1)^oo (7/(3^n) + 6/((n+3)(n+4))) = 21/2 + 1/4 = 43/4` .**