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
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\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_{k}&=&{\frac {1}{2}}\left({\frac {1}{1}}+{\frac {1}{2}}-{\frac {1}{k}}-{\frac {1}{k+1}}\right).\end{array}}}
Again, notice the pattern: each time there are exactly two surviving positive terms, and two surviving negative terms in each partial sum. 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):
This is the easier portion of the problem. Each term grows by a ratio of $1/3$, and it reverses sign. Thus, there is a common ratio $r=-1/3$. Also, the first term is $1/2$, so we can write the series as a geometric series:
$\sum _{n=0}^{\infty }\,{\frac {1}{2}}\left(-{\frac {1}{3}}\right)^{n}.$
Then, the series converges to the sum
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle S\,=\,{\frac {a_{0}}{1-r}}\,=\,{\frac {1/2}{1-(-1/3)}}\,=\,{\frac {1/2}{4/3}}\,=\,{\frac {3}{8}}.}

(b):
Using the technique in Foundations, we can rewrite the sequence as
$\sum _{n=1}^{\infty }\,{\frac {3}{(2n-1)(2n+1)}}\,=\,3\sum _{n=1}^{\infty }\,{\frac {1}{2}}\left({\frac {1}{2n-1}}-{\frac {1}{2n+1}}\right).$
Writing a few terms out, we find
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle S\,=\,{\frac {3}{2}}\left({\frac {1}{1}}-{\frac {1}{3}}+{\frac {1}{3}}-{\frac {1}{5}}+{\frac {1}{5}}-{\frac {1}{7}}+\cdots \right).}
Since only one positive term and one negative term survive, we have
$S_{k}\,=\,{\frac {3}{2}}\left(1-{\frac {1}{2k+1}}\right),$
so the series converges to the sum S, where
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle S\,=\,\lim _{k\rightarrow \infty }S_{k}\,=\,{\frac {3}{2}}.}

Final Answer:
The series in (a) converges to $3/8$, while the series in (b) converges to Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 3/2} .

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