009C Sample Midterm 3, Problem 3

Test if each the following series converges or diverges. Give reasons and clearly state if you are using any standard test.

(a) (6 points)      ${\displaystyle \sum _{n=1}^{\infty }}\,{\frac {n!}{(3n+1)!}}.$

(b) (6 points)      ${\displaystyle \sum _{n=2}^{\infty }}\,{\frac {\sqrt {n}}{n^{2}-3}}.$

Foundations:
Most of the time, if there are factorials in the terms of a series, you would use the
Ratio Test. Let $\sum _{k=1}^{\infty }a_{k}$ be a series. Then:
If $\lim _{k\rightarrow \infty }\left|{\frac {a_{k+1}}{a_{k}}}\right|=L<1$, the series is absolutely convergent (and therefore convergent).
If $\lim _{k\rightarrow \infty }\left|{\frac {a_{k+1}}{a_{k}}}\right|=L>1$ or $\lim _{k\rightarrow \infty }\left|{\frac {a_{k+1}}{a_{k}}}\right|=L=\infty$, the series is divergent.
If $\lim _{k\rightarrow \infty }\left|{\frac {a_{k+1}}{a_{k}}}\right|=L=1$, the Ratio Test is inconclusive.
This works well, as factorials cancel out many terms. For example,
${\frac {(n+1)!}{n!}}\,=\,n+1.$
On the other hand, something built mainly out of powers of $n$ may work well with the
Limit Comparison Test. Suppose $\sum _{k=1}^{\infty }a_{k}$ and $\sum _{k=1}^{\infty }b_{k}$ are series with positive terms. If $\lim _{k\rightarrow \infty }{\frac {a_{k}}{b_{k}}}=c$ where $0<c<\infty$, then either both series converge, or both series diverge.
In the case of a series built mainly out of powers, you would choose to compare it to a p-series.

 Solution:

(a):
As mentioned in Foundations, we should use the ratio test. Note that
${\begin{array}{rcl}\displaystyle {\left|{\frac {a_{n+1}}{a_{n}}}\right|}&=&\left|{\frac {\frac {(n+1)!}{(3(n+1)+1)!}}{\frac {n!}{(3n+1)!}}}\right|\\\\&=&\displaystyle {{\frac {(n+1)!}{n!}}\cdot {\frac {(3n+1)!}{(3n+4)!}}}\\\\&=&\displaystyle {\frac {n+1}{(3n+2)(3n+3)(3n+4)}}.\end{array}}$
Thus,
$\lim _{n\rightarrow \infty }\left|{\frac {a_{n+1}}{a_{n}}}\right|\ =\ 0\ <\ 1,$
so by the ratio test the series converges.

(b):
Here, we can use the limit comparison test. Let $a_{n}={\frac {\sqrt {n}}{n^{2}-3}}$, and let $b_{n}={\frac {1}{n^{3/2}}}.$ Notice that the terms of $a_{n}$ are all positive, and
${\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\frac {a_{n}}{b_{n}}}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\frac {\frac {\sqrt {n}}{n^{2}-3}}{\frac {1}{n^{3/2}}}}}\\\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\frac {n^{2}}{n^{2}-3}}}\\\\&=&1\,\,<\,\,\infty .\end{array}}$ Since $\sum _{n=2}^{\infty }{\frac {1}{n^{3/2}}}$  is a p-series with $p>1,$ it is convergent. By the limit comparison test, $\sum _{n=2}^{\infty }{\frac {\sqrt {n}}{n^{2}-3}}$  is convergent.

Final Answer:
Both series are convergent.

Return to Sample Exam