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,
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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  $p$-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
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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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