Question #266793

Determine for which value of Ξ±\alpha the given series converge absolutely, converge conditionally, or diverges.

  1. βˆ‘k=1∞((βˆ’1)\displaystyle\sum_{k=1}^ ∞ ((-1) k+1 * Ξ±k\alpha^k ) / k2
  2. βˆ‘k=1∞((Ξ±βˆ’1)k​\displaystyle\sum_{k=1}^ ∞ ((\alpha -1)^k ​ ) / k

For each of the following series, determine whether it is absolutely convergent, conditionally convergent, or divergent. Justify your answer.

  1. βˆ‘n=2∞(βˆ’1)n/(nlnn)\displaystyle\sum_{n=2}^ ∞ (-1)^n/(n ln n)
1
Expert's answer
2021-11-23T16:46:06-0500

1.

series diverges if

lim⁑kβ†’βˆžβˆ£ak∣=lim⁑kβ†’βˆžβˆ£Ξ±k/k2∣=βˆžβ€…β€ŠβŸΉβ€…β€Šβˆ£Ξ±βˆ£>1\displaystyle{\lim_{k\to \infin}}|a_k|=\displaystyle{\lim_{k\to \infin}}|\alpha^k/k^2|=\infin\implies |\alpha|>1


series converge absolutely if

lim⁑kβ†’βˆžβˆ£ak∣=lim⁑kβ†’βˆžβˆ£Ξ±k/k2∣=0β€…β€ŠβŸΉβ€…β€Šβˆ£Ξ±βˆ£<1\displaystyle{\lim_{k\to \infin}}|a_k|=\displaystyle{\lim_{k\to \infin}}|\alpha^k/k^2|=0\implies |\alpha|<1


2.

series diverges if

lim⁑kβ†’βˆžβˆ£ak∣=lim⁑kβ†’βˆžβˆ£(Ξ±βˆ’1)k/k∣=βˆžβ€…β€ŠβŸΉβ€…β€Šβˆ£Ξ±βˆ£>1\displaystyle{\lim_{k\to \infin}}|a_k|=\displaystyle{\lim_{k\to \infin}}|(\alpha-1)^k/k|=\infin\implies |\alpha|>1


series converge absolutely if

lim⁑kβ†’βˆžβˆ£ak∣=lim⁑kβ†’βˆžβˆ£(Ξ±βˆ’1)k/k∣=0β€…β€ŠβŸΉβ€…β€Šβˆ£Ξ±βˆ’1∣<1β€…β€ŠβŸΉβ€…β€Š0<Ξ±<1\displaystyle{\lim_{k\to \infin}}|a_k|=\displaystyle{\lim_{k\to \infin}}|(\alpha-1)^k/k|=0\implies |\alpha-1|<1\implies 0<\alpha<1 and 1<Ξ±<21<\alpha<2


3.

for alternating series:

series converges if

∣an∣|a_n| decreases monotonically, i. e. ∣an+1βˆ£β‰€βˆ£an∣|a_{n+1}|\le |a_n| and lim⁑nβ†’βˆžβˆ£an∣=0\displaystyle{\lim_{n\to \infin}}|a_n|=0


lim⁑nβ†’βˆžβˆ£an∣=lim⁑nβ†’βˆžβˆ£1/(nlnn)∣=0\displaystyle{\lim_{n\to \infin}}|a_n|=\displaystyle{\lim_{n\to \infin}}|1/(nlnn)|=0


1nlnnβ‰₯1(n+1)ln(n+1)\frac{1}{nlnn}\ge\frac{1}{(n+1)ln(n+1)}


so, series converge


for series of absolute values βˆ‘n=2∞1nlnn\displaystyle\sum_{n=2}^{\infin}\frac{1}{nlnn} :


∫2∞1xlnxdx=ln(lnx)∣2∞=∞\int_2^{\infin}\frac{1}{xlnx}dx=ln(lnx)|^{\infin}_2=\infin


so, βˆ‘n=2∞1nlnn\displaystyle\sum_{n=2}^{\infin}\frac{1}{nlnn} diverges, this means that βˆ‘n=2∞(βˆ’1)nnlnn\displaystyle\sum_{n=2}^{\infin}\frac{(-1)^n}{nlnn} converge conditionally





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