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)      Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \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
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {4}{(n-2)(n+1)}}} . First, consider the difference
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {1}{n-2}}-{\frac {1}{n-1}}}
If we combine this to a common denominator, we find
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 1/3,} or the reciprocal of the difference between the two constants. Thus
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {1}{(x+a)(x+b)}}} where Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle a<b,} we have
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\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
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\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
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \sum _{n=2}^{\infty }{\frac {1}{(n-1)(n+1)}}\,=\,\sum _{n=2}^{\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
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 1/3} , and it reverses sign. Thus, there is a common ratio Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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 (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_k \,=\,\frac{3}{2}\left( 1-\frac{1}{2k+1} \right),}
so the series converges to the sum S, where
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S\,=\,\lim_{k\rightarrow\infty} S_k\,=\,\frac{3}{2}.}

Final Answer:

The series in (a) converges to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 3/8} , while the series in (b) converges to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 3/2} .

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