009C Sample Final 3, Problem 2

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  $p={\frac {1}{2}}.$
Therefore, it diverges.
Hence, the series
$\sum _{n=1}^{\infty }{\frac {(-1)^{n}}{\sqrt {n}}}$
is not absolutely convergent.

(b)

Step 1:
For
$\sum _{n=1}^{\infty }{\frac {(-1)^{n}}{n}},$
we notice that this series is alternating.
Let  $b_{n}={\frac {1}{n}}.$
The sequence  $\{b_{n}\}$  is decreasing since
${\frac {1}{n+1}}<{\frac {1}{n}}$
for all  $n\geq 1.$
Also,
$\lim _{n\rightarrow \infty }b_{n}=\lim _{n\rightarrow \infty }{\frac {1}{n}}=0.$
Therefore, the series  $\sum _{n=1}^{\infty }{\frac {(-1)^{n}}{n}}$   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.

Final Answer:
(a)    not absolutely convergent
(b)    conditionally convergent

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