Question

How do you find `\sum _ { n = 1} ^ { \infty } \frac { 15^ { n } } { ( n + 1) 6^ { 2n + 1} }`?

Answer 1

`sum_(n=1)^oo15^n/((n+1)*6^(2n+1)) = 2/5ln(12/7)-1/6~~0.0489`

Explanation:

We will make use of the taylor series

`-sum_(n=1)^oox^n/n = ln(1-x)` for `|x|<1`

with that:

`sum_(n=1)^oo15^n/((n+1)*6^(2n+1)) = sum_(n=1)^oo15^n/((n+1)*6*36^n)`

`=1/6sum_(n=1)^oo1/(n+1)(15/36)^n`

`=1/6sum_(n=2)^oo(1/n)(5/12)^(n-1)`

`=1/6sum_(n=2)^oo(1/n)(5/12)^n(5/12)^(-1)`

`=12/5*1/6[-5/12+sum_(n=1)^oo(1/n)(5/12)^n]`

`=-1/6+2/5sum_(n=1)^oo(1/n)(5/12)^n`

`=-1/6-2/5[-sum_(n=1)^oo(1/n)(5/12)^n]`

`=-1/6-2/5ln(1-5/12)`

`=2/5ln(12/7)-1/6`

`~~0.0489`

Answer 2

`0.04893193362640811`

Explanation:

`sum_(n=1)^oo 15^n/((n+1)6^(2n+1))=6/15sum_(n=1)^oo((15/36)^(n+1))/(n+1)=2/5sum_(n=2)^oo x^n/n`

with `x = 15/36=5/12 < 1`

Now `2/5sum_(n=2)^oo x^n/n=2/5sum_(n=2)^oo int_0^xlambda^(n-1)dlambda = 2/5sum_(n=1)^oo int_0^xlambda^ndlambda` and also

`2/5sum_(n=1)^oo int_0^xlambda^ndlambda=2/5(int_0^x sum_(n=1)^oo lambda^n)dlambda` but `abs lambda < 1` so

`2/5(int_0^x sum_(n=1)^oo lambda^n)dlambda = 2/5int_0^x(-1/(lambda-1)-1)dlambda`

and finally

`sum_(n=1)^oo 15^n/((n+1)6^(2n+1))=2/5(-log(1-x)-x) = 2/5(-log(1-5/12)-5/12)=0.04893193362640811`