# 009C Sample Final 3, Problem 2

Consider the series

${\displaystyle \sum _{n=2}^{\infty }{\frac {(-1)^{n}}{\sqrt {n}}}.}$

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

Solution:

(a)

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

(b)

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

(a)    not absolutely convergent (by the  ${\displaystyle p}$-series test)