# 009C Sample Midterm 1, Problem 3 Detailed Solution

Determine whether the following series converges absolutely,

conditionally or whether it diverges.

$\sum _{n=1}^{\infty }{\frac {(-1)^{n}}{n}}$ Background Information:
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:

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