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

2 Answers
sente
Dec 12, 2016

#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#

Cesareo R.
Dec 12, 2016

#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#

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