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
You use the Cauchy test
if
if
if
take the limit
so it converge
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
- 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
#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
#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
# = 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