How do you use the ratio test to test the convergence of the series #∑ (n!)^2 / (kn)!# from n=1 to infinity?

Question

Isabella Ackerman · Accepted Answer

The series:

#sum_(n=1)^oo ((n!)^2)/((kn)!)#

is convergent for #k>=2# The ratio test states that a necessary condition for a series #sum_(n=1)^oo a_n# to converge is that: #L= lim_(n->oo) abs (a_(n+1)/a_n) <= 1# if # L < 1# the condition is also sufficient. In our case: #abs (a_(n+1)/a_n) = ((((n+1)!)^2)/((k(n+1))!))/(((n!)^2)/((kn)!)) = (((n+1)!)^2)/((n!)^2) ((kn)!)/ ((kn+k)!) = (n+1)^2/((kn+k) (kn+k-1)...(kn+1))# Now we have that: #lim_(n->oo) abs (a_(n+1)/a_n) = lim_(n->oo) (n+1)^2/(n+1) = lim_(n->oo) (n+1) = +oo# and the series is divergent. #lim_(n->oo) abs (a_(n+1)/a_n) = lim_(n->oo) (n+1)^2/((n+2)(n+1)) = 1# and the test is inconclusive, so we have to look at the series in more detail: #sum_(n=1)^oo ((n!)^2)/((2n)!) = (n(n-1)(n-2)...2*1)/ (2n(2n-1)...(n+1)# We can note that the numerator has #n# factors from #1# to #n# and the denominator has #n# factors from #n+1# to #2n#, so ordering them appropriately we have: # (n(n-1)(n-2)...2)/ (2n(2n-1)...(n+1)) = prod_(q=1)^n q/(n+q)# Now as #q <= n#, #q/(n+q) <= q/(q+q) = 1/2# So we have: # (n(n-1)(n-2)...2)/ (2n(2n-1)...(n+1)) <= (1/2)^n# And as: #sum_(n=1)^oo (1/2)^n = 1# is convergent, then also our series is convergent by direct comparison. #lim_(n->oo) abs (a_(n+1)/a_n) = lim_(n->oo) (n+1)^2/((n+1)(n+2)...(n+k)) = 0# and the series is convergent.

Ezra Adams · Answer

undefined To use the ratio test to test the convergence of the series ∑ (n!)^2 / (kn)! from n=1 to infinity, you compute the limit as n approaches infinity of the absolute value of the ratio of consecutive terms in the series, which is given by:

lim (n→∞) |[a_{n+1} / a_n]|

For the given series, the nth term is ((n!)^2) / ((kn)!). Therefore, the (n+1)th term is (((n+1)!)^2) / ((k(n+1))!).

Substitute these into the ratio test formula and simplify to find the limit. If the limit is less than 1, the series converges. If it is greater than 1, the series diverges. If it equals 1, the test is inconclusive.