Question

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?

Answer 1

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

Answer 2

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.