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. Ratio Test
Let  $\sum a_{n}$  be a series and  $L=\lim _{n\rightarrow \infty }{\bigg |}{\frac {a_{n+1}}{a_{n}}}{\bigg |}.$
Then,
If  $L<1,$  the series is absolutely convergent.
If  $L>1,$  the series is divergent.
If  $L=1,$  the test is inconclusive.
2. If a series absolutely converges, then it also converges.
3. Direct Comparison Test
Let  $\{a_{n}\}$  and  $\{b_{n}\}$  be positive sequences where  $a_{n}\leq b_{n}$
for all  $n\geq N$  for some  $N\geq 1.$
1. If  $\sum _{n=1}^{\infty }b_{n}$  converges, then  $\sum _{n=1}^{\infty }a_{n}$  converges.
2. If  $\sum _{n=1}^{\infty }a_{n}$  diverges, then  $\sum _{n=1}^{\infty }b_{n}$  diverges.

Solution:

(a)

Step 1:
First, we have
$\sum _{n=1}^{\infty }(-1)^{n}{\sqrt {\frac {1}{n}}}=\sum _{n=1}^{\infty }(-1)^{n}{\frac {1}{\sqrt {n}}}.$

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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