# 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?

Feb 9, 2016

You use the Cauchy test

${\sum}_{n = 1}^{\infty} {a}_{n} = {\sum}_{n = 1}^{\infty} \frac{{\left(4 n + 3\right)}^{n}}{{\left(n + 7\right)}^{2 n}}$

if ${\lim}_{n \to \infty} {a}_{n}^{\frac{1}{n}} < 1$ 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

${\lim}_{n \to \infty} \frac{4 + \frac{3}{n}}{n \left(1 + \frac{14}{n} + \frac{49}{n} ^ 2\right)} = 0$

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:

${\sum}_{n = 1}^{\infty} {\left(4 n + 3\right)}^{n} / {\left(n + 7\right)}^{2 n} = {\sum}_{n = 1}^{\infty} {\left(4 n + 3\right)}^{n} / {\left({\left(n + 7\right)}^{2}\right)}^{n} = {\sum}_{n = 1}^{\infty} {\left(\frac{4 n + 3}{{\left(n + 7\right)}^{2}}\right)}^{n}$

The root test states that for a series ${\sum}_{n = 1}^{\infty} {a}_{n}$,

• if ${\lim}_{n \to \infty} \sqrt[n]{\left\mid {a}_{n} \right\mid} < 1$, then ${\sum}_{n = 1}^{\infty} {a}_{n}$ converges
• if ${\lim}_{n \to \infty} \sqrt[n]{\left\mid {a}_{n} \right\mid} > 1$, then ${\sum}_{n = 1}^{\infty} {a}_{n}$ diverges
• if ${\lim}_{n \to \infty} \sqrt[n]{\left\mid {a}_{n} \right\mid} = 1$, 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.

${\lim}_{n \to \infty} \sqrt[n]{\left\mid {a}_{n} \right\mid} = {\lim}_{n \to \infty} \sqrt[n]{\left\mid {\left(\frac{4 n + 3}{{\left(n + 7\right)}^{2}}\right)}^{n} \right\mid}$

You can omit the absolute value since ${\left(\frac{4 n + 3}{{\left(n + 7\right)}^{2}}\right)}^{n} > 0$ is true for any $n > 0$. Thus, your calculation simplifies to:

${\lim}_{n \to \infty} \sqrt[n]{\left\mid {a}_{n} \right\mid} = {\lim}_{n \to \infty} \frac{4 n + 3}{{\left(n + 7\right)}^{2}}$

$= {\lim}_{n \to \infty} \frac{4 n + 3}{{n}^{2} + 14 n + 49}$

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

$= {\lim}_{n \to \infty} \frac{{n}^{2} \left(\frac{4}{n} + \frac{3}{n} ^ 2\right)}{{n}^{2} \left(1 + \frac{14}{n} + \frac{49}{n} ^ 2\right)}$

$= {\lim}_{n \to \infty} \frac{\cancel{{n}^{2}} \left(\frac{4}{n} + \frac{3}{n} ^ 2\right)}{\cancel{{n}^{2}} \left(1 + \frac{14}{n} + \frac{49}{n} ^ 2\right)}$

$= {\lim}_{n \to \infty} \frac{\frac{4}{n} + \frac{3}{n} ^ 2}{1 + \frac{14}{n} + \frac{49}{n} ^ 2}$

... take the limit...

$= \frac{0 + 0}{1 + 0 + 0}$

$= 0$

As ${\lim}_{n \to \infty} \sqrt[n]{\left\mid {a}_{n} \right\mid} = 0 < 1$, according to the root test, the series converges.