How do you apply the ratio test to determine if #sum (1*3*5* * * (2n-1))/(1*4*7* * * (3n-2))# from #n=[1,oo)# is convergent to divergent?

Question

Paisley Johnson · Accepted Answer

the series is convergent.

We can apply d'Alembert's ratio test: Suppose that; # S=sum_(r=1)^oo a_n \ \ #, and #\ \ L=lim_(n rarr oo) |a_(n+1)/a_n| # Then if L < 1 then the series converges absolutely; if L > 1 then the series is divergent; if L = 1 or the limit fails to exist the test is inconclusive. So our series is; # S=sum_(n=1)^oo { 1*3*5 * ... * (2n-1) } / (1*4*7* ... * (3n-2) } # So our test limit is: # L = lim_(n rarr oo) | { { 1*3*5 * ... * (2n-1)* (2(n+1)-1) } / (1*4*7* ... * (3n-2)* (3(n+1)-2) } } / { { 1*3*5 * ... * (2n-1) } / (1*4*7* ... * (3n-2) } } | # # L = lim_(n rarr oo) | { 1*3 * ... * (2n-1) (2(n+1)-1) } / { 1*4* ... * (3n-2) (3(n+1)-2) } * {1*4*7* ... * (3n-2) } / { 1*3*5 * ... * (2n-1) } | # # L = lim_(n rarr oo) | { color(red)cancel(1*3 * ... * (2n-1)) (2(n+1)-1) } / { color(blue)cancel(1*4* ... * (3n-2)) (3(n+1)-2) } * {color(blue)cancel(1*4*7* ... * (3n-2)) } / { color(red)cancel(1*3*5 * ... * (2n-1)) } | # # L = lim_(n rarr oo) | (2(n+1)-1) / (3(n+1)-2) | # # \ \ \ = lim_(n rarr oo) | (2n+1)/(3n+1) | # # \ \ \ = lim_(n rarr oo) | (2n+1)/(3n+1) *(1/n)/(1/n)| # # \ \ \ = lim_(n rarr oo) | (2+1/n)/(3+1/n) | # # \ \ \ = 2/3 # And so the series is convergent.

Christopher Agresta · Answer

undefined To apply the ratio test to determine the convergence or divergence of the series $ \sum_{n=1}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n-1)}{1 \cdot 4 \cdot 7 \cdot \ldots \cdot (3n-2)} $, follow these steps:

1. Compute the ratio of consecutive terms:
   $$ r = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| $$
   where $ a_n $ represents the $ n $th term of the series.

2. Substitute the expressions for $ a_n $ and $ a_{n+1} $ into the ratio:
   $$ r = \lim_{n \to \infty} \left| \frac{(2(n+1)-1)}{(3(n+1)-2)} \cdot \frac{(3n-2)}{(2n-1)} \right| $$

3. Simplify the expression and compute the limit.

4. If the limit $ r $ is less than 1, the series converges. If $ r $ is greater than 1 or the limit does not exist, the series diverges.

5. Based on the result obtained in step 4, conclude whether the series $ \sum_{n=1}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n-1)}{1 \cdot 4 \cdot 7 \cdot \ldots \cdot (3n-2)} $ is convergent or divergent.