009C Sample Final 3, Problem 2

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Consider the series

$\sum _{n=2}^{\infty }{\frac {(-1)^{n}}{\sqrt {n}}}.$ (a) Test if the series converges absolutely. Give reasons for your answer.

(b) Test if the series converges conditionally. Give reasons for your answer.

Foundations:
1. A series  $\sum a_{n}$ is absolutely convergent if
the series  $\sum |a_{n}|$ converges.
2. A series  $\sum a_{n}$ is conditionally convergent if
the series  $\sum |a_{n}|$ diverges and the series  $\sum a_{n}$ converges.

Solution:

(a)

Step 1:
First, we take the absolute value of the terms in the original series.
Let  $a_{n}={\frac {(-1)^{n}}{\sqrt {n}}}.$ Therefore,
${\begin{array}{rcl}\displaystyle {\sum _{n=1}^{\infty }|a_{n}|}&=&\displaystyle {\sum _{n=1}^{\infty }{\bigg |}{\frac {(-1)^{n}}{\sqrt {n}}}{\bigg |}}\\&&\\&=&\displaystyle {\sum _{n=1}^{\infty }{\frac {1}{\sqrt {n}}}.}\end{array}}$ Step 2:
This series is a  $p$ -series with  $p=1/2.$ Therefore, it diverges.
Hence, the series
$\sum _{n=1}^{\infty }{\frac {(-1)^{n}}{\sqrt {n}}}$ is not absolutely convergent.

(b)

Step 1:
For
$\sum _{n=1}^{\infty }{\frac {(-1)^{n}}{\sqrt {n}}},$ we notice that this series is alternating.
Let  $b_{n}={\frac {1}{\sqrt {n}}}.$ First, we have
${\frac {1}{\sqrt {n}}}\geq 0$ for all  $n\geq 1.$ The sequence  $\{b_{n}\}$ is decreasing since
${\frac {1}{\sqrt {n+1}}}<{\frac {1}{\sqrt {n}}}$ for all  $n\geq 1.$ Also,
$\lim _{n\rightarrow \infty }b_{n}=\lim _{n\rightarrow \infty }{\frac {1}{\sqrt {n}}}=0.$ Therefore, the series  $\sum _{n=1}^{\infty }{\frac {(-1)^{n}}{\sqrt {n}}}$ converges
by the Alternating Series Test.
Step 2:
Since the series  $\sum _{n=1}^{\infty }{\frac {(-1)^{n}}{\sqrt {n}}}$ is not absolutely convergent but convergent,
this series is conditionally convergent.

(a)    not absolutely convergent (by the  $p$ -series test)