009C Sample Midterm 1, Problem 3

Determine whether the following series converges absolutely, conditionally or whether it diverges.

Be sure to justify your answers!

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

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:

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 $p=1$).
So, 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 is conditionally converges.
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 4:
Since the series $\sum _{n=1}^{\infty }{\frac {(-1)^{n}}{n}}$ is not absolutely convergent but convergent,
this series is conditionally convergent.

Final Answer:
Conditionally convergent

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