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

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: `abs(a_{k+1}/a_k)`;
3. Try to compute the limit of this ratio: `lim_{k\to\infty}abs(a_{k+1}/a_k)`;

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*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\to\infty} 3/(k+2)=0`, and thus the series converges.