Notes PDF
Introductory Problems
Determine in the series a1 + a2 + a3 + … converges or diverges.
1. an= (2⋅4⋅ … ⋅(2n))/(2n)!
Answer
1. The series converges.
Solution
1. Factorials suggest the ratio test. The only trick is to simplify the numerator of an by factoring a 2 out of factor and grouping them together, so 2⋅4⋅ … ⋅(2n) = 2nn! Then after simplification an+1/an= (2n+1/2n)⋅((n+1)!/(n!))⋅((2n)!/(2n+2)!) = (2(n+1)((2n)!))/((2n+2)(2n+1) ((2n)!)) = 1/(2n+1) This → 0 as n → ∞, so the series converges by the Ratio Test.
2. an= (nn)/(2n(2))
Answer
1. The series converges.
Solution
1. The presence of n in exponents suggests the Root Test. The trick is to recognize the denominator is 2n(2)= (2n)n, so an can be rewritten as (nn)/((2n)n)= (n/2n)n Then (an)1/n= n/2n. Replacing n with x and applying l’Hopital’s rule, we see: limn → ∞(an)1/n = 0 so the series converges by the Root Test.
3. an= ((3n)1)/(n1(2n)1)
Answer
1. The series diverges.
Solution
1. Factorials suggest the ratio test. Grouping like factors together we find an+1/an= ((3n+3)!/(3n)!)⋅(n!/(n+1)!)⋅((2n)!/(2n+2)!) Canceling common factors gives an+1/an= ((3n+3) (3n+2)(3n+1))/((n+1)(2n+1)(2n+2)) = (27n3 + quadratic and lower terms)/(4n3 + quadratic and lower terms) As n → ∞, the cubic terms dominate, so limn → ∞an+1/an= 27/4. The series diverges by the Ratio Test.