Does the series #sum_{n=1}^oo (5n)^(3n)/(5^n+3)^n# diverge or converge?

2 Answers
Feb 20, 2018

In order to find the result, you must use the Ratio Test.

Explanation:

The ratio test says that :
#sum_(n=1)^∞ a_n# converges only and only
#|a_(k+1)/a_k| < 1#, and that it diverges when it's #>1#. When it's equal, it's inconclusive.

So, you'll reach
#lim_(n->∞)|(((5n+5)^(3n+3))/(5^(n+1)+3))^(n+1) * ((5^n+3)^3/(5n))^n|#.

Take each term of the two different terms and use the natural logarithm and L'Hôspital's rule.

Feb 21, 2018

See below.

Explanation:

#(5^(3n)n^(3n))/(5^n+3)^n < (5^(3n)n^(3n))/(5^n)^n = n^(3n)/(5^((n^2)-3n))#

now

#n^(3n) = 5^(3n(logn/log5))# then

#n^(3n)/(5^(n^2-3n)) = 1/(5^(n^2-3n-3n(logn/log5)))#

now for #n > 8, n in ZZ# we have #n^2-3n(1+(logn/log5))> n#

and then for #n > 8#

#a_n = ((5n)^(3n))/(5^n+3)^n < 1/5^n#

hence

#sum_(n=1)^oo ((5n)^(3n))/(5^n+3)^n # converges because

#sum_(k=1)^oo 1/5^k# converges.