009C Sample Midterm 3, Problem 3 Detailed Solution

Test if each the following series converges or diverges.

Give reasons and clearly state if you are using any standard test.

(a)  ${\displaystyle \sum _{n=1}^{\infty }}\,{\frac {n!}{(3n+1)!}}$

(b)  ${\displaystyle \sum _{n=2}^{\infty }}\,{\frac {\sqrt {n}}{n^{2}-3}}$

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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.

Final Answer:
conditionally convergent (by the p-test and the Alternating Series Test)

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