How do you use the ratio test to test the convergence of the series #∑ ((4n+3)^n) / ((n+7)^(2n))# from n=1 to infinity?

2 Answers
Feb 9, 2016

You use the Cauchy test

#sum_(n=1)^oo a_n = sum_(n=1)^(oo)((4n+3)^n) / ((n+7)^(2n))#

if #lim_(n->oo)a_n^(1/n) < 1# then the series converge

if #> 1# then it diverge

if #= 1# you can't conclude

#(((4n+3)^n) / ((n+7)^(2n)))^(1/n) = (4n+3)/(n+7)^2 = (4n+3)/(n^2+14n+49) = (4+3/n)/(n(1+14/n+49/n^2)#

take the limit

#lim_(n->oo) (4+3/n)/(n(1+14/n+49/n^2)) = 0#

so it converge

Feb 9, 2016

According to the root test, the series converges.

Explanation:

I know that you have asked for the ratio test to test the convergence.

However, in this case, I would strongly recommend to do the root test instead.

You can transform your series as follows:

# sum_(n=1)^(oo) (4n +3)^n/(n+7)^(2n) = sum_(n=1)^(oo) (4n+3)^n/((n+7)^2)^n = sum_(n=1)^(oo) ((4n+3)/((n+7)^2))^n#

The root test states that for a series #sum_(n=1)^(oo) a_n#,

  • if #lim_(n->oo) root(n)(abs(a_n)) < 1#, then #sum_(n=1)^(oo) a_n# converges
  • if #lim_(n->oo) root(n)(abs(a_n)) > 1#, then #sum_(n=1)^(oo) a_n# diverges
  • if #lim_(n->oo) root(n)(abs(a_n)) = 1#, then you can't tell with a root test.

As you can transform your #a_n# as an expression taken to the power of #n#, the root test is really well suited.

#lim_(n->oo) root(n)(abs(a_n)) = lim_(n->oo) root(n)(abs( ((4n+3)/((n+7)^2))^n ))#

You can omit the absolute value since #((4n+3)/((n+7)^2))^n > 0# is true for any #n > 0#. Thus, your calculation simplifies to:

#lim_(n->oo) root(n)(abs(a_n)) = lim_(n->oo) (4n+3)/((n+7)^2)#

# = lim_(n->oo) (4n+3)/(n^2 + 14n + 49)#

... the highest power of #n# in your fraction is #n^2#. Factor #n^2# in the numerator and denominator...

# = lim_(n->oo) (n^2 (4/n +3/n^2 ))/(n^2( 1 + 14/n + 49/n^2))#

# = lim_(n->oo) (cancel(n^2) (4/n +3/n^2 ))/(cancel(n^2)( 1 + 14/n + 49/n^2))#

# = lim_(n->oo) ( 4/n +3/n^2 )/( 1 + 14/n + 49/n^2)#

... take the limit...

# = ( 0 + 0 )/( 1 + 0 + 0)#

# = 0#

As #lim_(n->oo) root(n)(abs(a_n)) = 0 < 1#, according to the root test, the series converges.