Series practice
When answering the following, keep in mind series and series tests that we have discussed …
nth term divergence test
telescoping
geometric
integral test
p-series
direct comparison
limit comparison
alternating series test
absolute convergence
root test
ratio test
Tasks
Determine convergence or divergence of the following series.
1. \( \displaystyle\sum_{n=1}^{\infty} 2^{1/n} \)
hint 1 how can we test for divergence?
hint 2 \( \displaystyle\lim_{n \to \infty} 2^{1/n} = 2^{\lim_{n \to \infty} 1/n } \)
2. \( \displaystyle\sum_{n=1}^{\infty} \dfrac{n!}{(2n)!} \)
hint 1 \( a_{n+1} = \dfrac{(n+1)!}{(2(n+1))!} = \dfrac{(n+1)!}{(2n+2)!} \)
hint 2 \( a_{n+1} = \dfrac{(n+1) (n)!}{ (2n+2)(2n+1)(2n)! } \)
3. \( \displaystyle\sum_{n=1}^{\infty} \dfrac{ (-1)^{n-1} }{\sqrt{n}} \)
hint 1 alternating series test?
hint 2 \( b_n = \dfrac{1}{\sqrt{n} }\)
hint 3 \( b_n = \dfrac{1}{\sqrt{n}} \) is positive, decreasing, and \( \displaystyle\lim_{n \to \infty} b_n = 0 \)
4. \( \displaystyle\sum_{n=2}^{\infty} \dfrac{1}{n (\ln n)^3} \)
hint 1 \( f(x) = \dfrac{1}{x (\ln x)^3} \) is positive, decreasing, and continuous
hint 2 \( u = \ln x, du = \dfrac{1}{x} dx \)
hint 3 \( \displaystyle\int \dfrac{1}{u^3} \,du \)
5. \( \displaystyle\sum_{n=1}^{\infty} 4^{-2n+1} \)
hint 1 exponential rules?
hint 2 \( 4^{-2n + 1} = 4^{-2n} \cdot 4 \)
hint 3 \( 4^{-2n} \cdot 4 = \dfrac{4}{4^{2n}} = \dfrac{4}{16^n} = 4 \cdot \left( \dfrac{1}{16} \right)^n \)
Answers (mixed up)
diverges by the nth term divergence test
series converges by the alternating series test
geometric series with \( r = \dfrac{1}{16}, \left| \dfrac{1}{16} \right| \lt 1 \), so series converges
ratio test equals 0, so series converges
series converges by integral test