How do you test the series #Sigma (100^n(n!)^3)/((3n)!)# from #n=[1,oo)# by the ratio test?

Question

Ezekiel Alexander · Accepted Answer

The series:

#sum_(n=1)^oo((100^n(n!)^3)/((3n)!))#

is divergent.

The ratio test states that given the series: #sum_(n=1)^oo a_n# and the limit: #L = lim_(n->oo) abs (a_(n+1)/a_n)# the series is absolutely convergent if #L < 1# and divergent if # L > 1#. Let's calculate the ratio for our series: # abs (a_(n+1)/a_n) = abs ( ( (100^(n+1)((n+1)!)^3)/((3(n+1))!)) /((100^n(n!)^3)/((3n)!))) # # abs (a_(n+1)/a_n) = ( 100^(n+1)/100^n) (((n+1)!)^3/ (n!)^3) ( ((3n)!)/((3n+3)!))# # abs (a_(n+1)/a_n) = 100 (((n+1)!)/(n!))^3 ( ((3n)!) / ( (3n+3)(3n+2)(3n+1) (3n)!))# # abs (a_(n+1)/a_n) = (100 (n+1)^3)/( (3n+3)(3n+2)(3n+1))# So: #lim_(n->oo) abs (a_(n+1)/a_n) = 100/27 > 1# and the series is divergent.

Axel Alvarez · Answer

undefined To test the series Σ (100^n(n!)^3)/((3n)!) using the ratio test:

1. Compute the ratio of consecutive terms:
   R = |a_{n+1} / a_n| where a_n represents the nth term of the series.

2. Evaluate the limit of R as n approaches infinity.

3. Determine convergence based on the limit of R:
   - If the limit is less than 1, the series converges.
   - If the limit is greater than 1 or does not exist, the series diverges.
   - If the limit equals 1, the test is inconclusive.

Now, apply the ratio test to the given series:

a_n = (100^n(n!)^3) / ((3n)!)

a_{n+1} = (100^(n+1)((n+1)!)^3) / ((3(n+1))!)

R = |(100^(n+1)((n+1)!)^3) / ((3(n+1))!)| / |(100^n(n!)^3) / ((3n)!)|
   
R = |(100^(n+1)((n+1)!)^3)(3n)!| / |(100^n(n!)^3)(3(n+1))!)|
   
R = |(100(n+1))^3| / |(3(n+1))(3(n+1)-1)(3(n+1)-2)|

R = (100(n+1))^3 / (3(n+1))(3n+2)(3n+1)(3n)

Now, take the limit of R as n approaches infinity:

lim (n→∞) R = lim (n→∞) [(100(n+1))^3 / (3(n+1))(3n+2)(3n+1)(3n)]

As n approaches infinity, the terms involving n in the denominator dominate, leading to a limit of 0.

Since the limit is less than 1, the series Σ (100^n(n!)^3)/((3n)!) converges by the ratio test.