Question #255792

Determine whether the following series converge, converge absolutely, converge conditionally, or diverge.

  1. k=1\displaystyle\sum_{k=1}^ ∞ (k+1/k)(√k+1/√k)-3
  2. k=1\displaystyle\sum_{k=1}^ ∞ (cos(kπ/2)/(2k)π(cos(k\pi/2)/(2k)\pi
  3. k=1\displaystyle\sum_{k=1}^ ∞(-1)k+1 (k1)!(k+1)!/(2k)!(k-1)!(k+1)!/(2k)!
  4. k=1\displaystyle\sum_{k=1}^ ∞ (1.3.5...(2k1))/(1.4.4.7...(3k2)))(1.3.5...(2k-1))/(1.4.4.7...(3k-2)))
1
Expert's answer
2021-10-25T16:23:46-0400

1.

11xpdx\int^{\infin}_1\frac{1}{x^p}dx is convergent if and only if p>1p>1


limk(k+1/k)3(k)3=10\displaystyle{\lim_{k\to \infin}}(\sqrt{k}+1/\sqrt k)^{-3}(\sqrt k)^3=1\neq 0

Since series (1/k3/2)\sum(1/k^{3/2}) converge, series k=1\displaystyle\sum_{k=1}^ ∞ (k+1/k)(√k+1/√k)-3 converge also.


2.

k=1\displaystyle\sum_{k=1}^ ∞ (cos(kπ/2)/(2k)π(cos(k\pi/2)/(2k)\pi=k=1\displaystyle\sum_{k=1}^ ∞ (1)kπ4k\frac{(-1)^{k}\pi}{4k}


Since k=1\displaystyle\sum_{k=1}^ ∞ 1k\frac{1}{k} diverge and limk(cos(kπ/2)/(2k)π=0\displaystyle{\lim_{k\to \infin}}|(cos(k\pi/2)/(2k)\pi|=0 , the series conditionally convergent.


3.

limk(k+2)!k!(2k+2)!(2k)!(k+1)!(k1)!=limk(k+2)k(2k+1)(2k+2)=1/4<1\displaystyle{\lim_{k\to \infin}}\frac{(k+2)!k!}{(2k+2)!}\cdot\frac{(2k)!}{(k+1)!(k-1)!}=\displaystyle{\lim_{k\to \infin}}\frac{(k+2)k}{(2k+1)(2k+2)}=1/4<1


Since k=1ak\displaystyle\sum_{k=1}^ ∞|a_k| converge, the series absolutely converge.


4.

limkak+1ak=limk2(k+1)13(k+1)2=2/3<1\displaystyle{\lim_{k\to \infin}}\frac{a_{k+1}}{a_k}=\displaystyle{\lim_{k\to \infin}}\frac{2(k+1)-1}{3(k+1)-2}=2/3<1


The series converge.


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