009C Sample Midterm 3, Problem 2

For each the following series find the sum, if it converges. If you think it diverges, explain why.

(a) (6 points)      ${\frac {1}{2}}-{\frac {1}{2\cdot 3}}+{\frac {1}{2\cdot 3^{2}}}-{\frac {1}{2\cdot 3^{3}}}+{\frac {1}{2\cdot 3^{4}}}-{\frac {1}{2\cdot 3^{5}}}+\cdots .$

(b) (6 points)      $\sum _{n=1}^{\infty }\,{\frac {3}{(2n-1)(2n+1)}}.$

Foundations:
One of the important series to know is the Geometric series. These are series with a common ratio $r$ between adjacent terms which are usually written
$\sum _{k=0}^{\infty }a_{0}r^{k}.$
These are convergent if $|r|<1$, and divergent if $|r|\geq 1$. If it is convergent, we can find the sum by the formula
$S={\frac {a_{0}}{1-r}},$
where $a_{0}$ is the first term in the series (if the index starts at $k=2$ or $k=6$, then "$a_{0}$" is actually the first term $a_{2}$ or $a_{6}$, respectively).
Another common type of series to evaluate is a telescoping series, where the telescoping better describes the partial sums, denoted $S_{k}$. Most of the time, they are presented as a fraction which requires partial fraction decomposition.
This can be accomplished fairly quickly via a shortcut when the factors in the denominator are linear and share the same coefficient on $n$.
Example. Suppose we wish to decompose the fraction ${\frac {4}{(n-2)(n+1)}}$. First, consider the difference
${\frac {1}{n-2}}-{\frac {1}{n-1}}$
If we combine this to a common denominator, we find
${\frac {1}{n-2}}\cdot {\frac {n+1}{n+1}}-{\frac {1}{n+1}}\cdot {\frac {n-2}{n-2}}\,=\,{\frac {n+1-(n-2)}{(n-2)(n+1)}}\ =\ {\frac {3}{(n-2)(n+1)}}.$
To have a 1 in the numerator, we would just multiply by $1/3,$ or the reciprocal of the difference between the two constants. Thus
${\begin{array}{rcl}{\displaystyle {\frac {4}{(n-2)(n+1)}}}&=&{\displaystyle 4\cdot {\frac {1}{(n-2)(n+1)}}}\\\\&=&4\cdot {\displaystyle {\frac {1}{3}}\left({\frac {1}{n-2}}-{\frac {1}{n+1}}\right).}\end{array}}$
Notice the pattern: for any fraction of the form${\frac {1}{(x+a)(x+b)}}$ where$a<b,$ we have
${\displaystyle {\frac {1}{(x+a)(x+b)}}\,=\,{\frac {1}{b-a}}\left({\frac {1}{x+a}}-{\frac {1}{x+b}}\right).}$
In this manner, we can quickly find that
${\displaystyle {\frac {1}{n^{2}-25}}\,=\,{\frac {1}{(n-5)(n+5)}}\,=\,{\frac {1}{5-(-5)}}\left({\frac {1}{n-5}}-{\frac {1}{n+5}}\right)\,=\,{\frac {1}{10}}\left({\frac {1}{n-5}}-{\frac {1}{n+5}}\right)}$
As per the so-called telescoping, consider the series defined by
$\sum _{n=2}^{\infty }{\frac {1}{n^{2}-1}}.$
Using the technique above, we can rewrite the series as
$\sum _{n=2}^{\infty }{\frac {1}{(n-1)(n+1)}}\,=\,\sum _{n=1}^{\infty }{\frac {1}{2}}\left({\frac {1}{n-1}}-{\frac {1}{n+1}}\right).$
This means that
${\begin{array}{cclcl}S_{2}&=&{\frac {1}{2}}\left({\frac {1}{1}}-{\frac {1}{3}}\right)\\S_{3}&=&{\frac {1}{2}}\left({\frac {1}{1}}-{\frac {1}{3}}+{\frac {1}{2}}-{\frac {1}{4}}\right)&=&{\frac {1}{2}}\left({\frac {1}{1}}+{\frac {1}{2}}-{\frac {1}{3}}-{\frac {1}{4}}\right)\\S_{4}&=&{\frac {1}{2}}\left({\frac {1}{1}}-{\frac {1}{3}}+{\frac {1}{2}}-{\frac {1}{4}}+{\frac {1}{3}}-{\frac {1}{5}}\right)&=&{\frac {1}{2}}\left({\frac {1}{1}}+{\frac {1}{2}}-{\frac {1}{4}}-{\frac {1}{5}}\right)\\S_{5}&=&{\frac {1}{2}}\left({\frac {1}{1}}-{\frac {1}{3}}+{\frac {1}{2}}-{\frac {1}{4}}+{\frac {1}{3}}-{\frac {1}{5}}+{\frac {1}{4}}-{\frac {1}{6}}\right)&=&{\frac {1}{2}}\left({\frac {1}{1}}+{\frac {1}{2}}-{\frac {1}{5}}-{\frac {1}{6}}\right)\\\\\vdots &\vdots &\vdots \\S_{n}&=&{\frac {1}{2}}\left({\frac {1}{1}}+{\frac {1}{2}}-{\frac {1}{n}}-{\frac {1}{n+1}}\right).\end{array}}$
Notice the pattern - each time there are exactly two surviving positive terms, and two surviving negative terms. This is exactly the difference between the two factors in the denominator. If we then take the limit, we find
$\sum _{n=2}^{\infty }{\frac {1}{n^{2}-1}}\,=\,\lim _{n\rightarrow \infty }S_{n}\,=\,\lim _{n\rightarrow \infty }{\frac {1}{2}}\left({\frac {1}{1}}+{\frac {1}{2}}-{\frac {1}{n}}-{\frac {1}{n+1}}\right)\,=\,{\frac {3}{4}}.$

 Solution:

(a):

(b):

Final Answer:

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