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

Oct 14, 2015

#### Answer:

The series converges absolutely.

#### Explanation:

The ratio test states the following:

1. Consider two consecutive terms ${a}_{k}$ and ${a}_{k + 1}$;
2. Divide the latter by the former and consider the absolute value: $\left\mid {a}_{k + 1} / {a}_{k} \right\mid$;
3. Try to compute the limit of this ratio: ${\lim}_{k \setminus \to \setminus \infty} \left\mid {a}_{k + 1} / {a}_{k} \right\mid$;

THEN, if the limit exists:

• If it's bigger then $1$ (strictly) , the series does not converge;
• If it's smaller then $1$ (strictly), the series converges absolutely;
• If it equals one, the test is inconclusive, because there exist both convergent and divergent series that satisfy this case.

In your case, a_k=3^k/((k+1)!), so

a_{k+1} = 3^{k+1}/((k+2)!)

Before dividing, it is useful to consider that:

• ${3}^{k + 1} = 3 \cdot {3}^{k}$,
• (k+2)! =(k+2)(k+1)!, and that
• dividing by a fraction means to multiply for the inverse of that fraction.

Now we can divide:

a_{k+1}/a_k= {3*3^k}/{(k+2)(k+1)!} * {(k+1)!}/3^k

We can simplify a lot of stuff:

 {3*color(red)(cancel(3^k))}/{(k+2)color(blue)(cancel((k+1))!)} * {color(blue)cancel((k+1)!)}/color(red)(cancel(3^k))

Now we can easily take the limit:

${\lim}_{k \setminus \to \setminus \infty} \frac{3}{k + 2} = 0$, and thus the series converges.