# 009C Sample Midterm 1, Problem 3 Detailed Solution

Determine whether the following series converges absolutely,

conditionally or whether it diverges.

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

Background Information:
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:

Step 1:
First, we take the absolute value of the terms in the original series.
Let  ${\displaystyle a_{n}={\frac {(-1)^{n}}{n}}.}$
Therefore,
${\displaystyle {\begin{array}{rcl}\displaystyle {\sum _{n=1}^{\infty }|a_{n}|}&=&\displaystyle {\sum _{n=1}^{\infty }{\bigg |}{\frac {(-1)^{n}}{n}}{\bigg |}}\\&&\\&=&\displaystyle {\sum _{n=1}^{\infty }{\frac {1}{n}}.}\end{array}}}$
Step 2:
This series is the harmonic series (or  ${\displaystyle p}$-series with  ${\displaystyle p=1}$ ).
Thus, it diverges. Hence, the series
${\displaystyle \sum _{n=1}^{\infty }{\frac {(-1)^{n}}{n}}}$
is not absolutely convergent.
Step 3:
Now, we need to look back at the original series to see
if it conditionally converges.
For
${\displaystyle \sum _{n=1}^{\infty }{\frac {(-1)^{n}}{n}},}$
we notice that this series is alternating.
Let  ${\displaystyle b_{n}={\frac {1}{n}}.}$
First, we have
${\displaystyle {\frac {1}{n}}\geq 0}$
for all  ${\displaystyle n\geq 1.}$
The sequence  ${\displaystyle \{b_{n}\}}$  is decreasing since
${\displaystyle {\frac {1}{n+1}}<{\frac {1}{n}}}$
for all  ${\displaystyle n\geq 1.}$
Also,
${\displaystyle \lim _{n\rightarrow \infty }b_{n}=\lim _{n\rightarrow \infty }{\frac {1}{n}}=0.}$
Therefore, the series  ${\displaystyle \sum _{n=1}^{\infty }{\frac {(-1)^{n}}{n}}}$   converges
by the Alternating Series Test.
Step 4:
Since the series
${\displaystyle \sum _{n=1}^{\infty }{\frac {(-1)^{n}}{n}}}$
converges but does not converge absolutely,
the series converges conditionally.