Consider two series \sum ↙\stackrel{\infty}{n = 1}⁡ (−1) ^{n−1\:}a_{n} and \sum ↙\stackrel{\infty}{n = 2}⁡ (−1) ^{n−1\:}b_{n}, where a_{n} = {1}/{\sqrt{n}},\:b_{n} = \frac{x^{n}}{n (n−1) } 0 < x < 1. Then:

Correct Answer is : both series are convergent

\sum ↙\stackrel{\infty}{n = 1}⁡ (−1) ^{n−1\:}a_{n} is convergent but \sum ↙\stackrel{\infty}{n = 2}⁡ (−1) ^{n−1\:}b_{n} is divergent.

\sum ↙\stackrel{\infty}{n = 1}⁡ (−1) ^{n−1\:}a_{n} is divergent but \sum ↙\stackrel{\infty}{n = 2}⁡ (−1) ^{n−1\:}b_{n} is convergent

AAI ATC Junior Executive 2018 Shift 3 Paper

Question : 88 of 120

Marks: +1, -0

Consider two series

∑ ↙{n = 1}↖{∞}⁡ (−1) ^{n−1\:}a_{n}

and

∑ ↙{n = 2}↖{∞}⁡ (−1) ^{n−1\:}b_{n}

, where

a_{n} = \;{1}/{{√{n}}},\:b_{n} = \;{x^{n}}/{n (n−1) }

0 < x < 1. Then:

$∑ ↙{n = 1}↖{∞}⁡ (−1) ^{n−1\:}a_{n}$ is convergent but $∑ ↙{n = 2}↖{∞}⁡ (−1) ^{n−1\:}b_{n}$ is divergent.
both series are convergent
$∑ ↙{n = 1}↖{∞}⁡ (−1) ^{n−1\:}a_{n}$ is divergent but $∑ ↙{n = 2}↖{∞}⁡ (−1) ^{n−1\:}b_{n}$ is convergent
both series are divergent.

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