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

Question

Charles Anctil · Accepted Answer

The series converges absolutely.

The ratio test states the following: THEN, if the limit exists: 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: 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 o\infty} 3/(k+2)=0#, and thus the series converges.

Luke Alvarez · Answer

undefined To use the ratio test to test the convergence of the series $ \sum \frac{3^k}{(k+1)!} $ from $ n = 1 $ to $ \infty $, you calculate the limit:

$$ L = \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| $$

where $ a_k = \frac{3^k}{(k+1)!} $.

Substitute $ a_{k+1} $ and $ a_k $ into the limit expression and simplify:

$$ L = \lim_{k \to \infty} \left| \frac{\frac{3^{k+1}}{(k+2)!}}{\frac{3^k}{(k+1)!}} \right| $$

$$ L = \lim_{k \to \infty} \left| \frac{3^{k+1}(k+1)!}{(k+2)! \cdot 3^k} \right| $$

$$ L = \lim_{k \to \infty} \left| \frac{3 \cdot (k+1)!}{(k+2)(k+1)!} \right| $$

$$ L = \lim_{k \to \infty} \left| \frac{3}{k+2} \right| $$

$$ L = 0 $$

Since the limit $ L $ is less than 1, by the ratio test, the series $ \sum \frac{3^k}{(k+1)!} $ converges absolutely.