| Foundations:
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One of the important series to know is the Geometric series. These are series with a common ratio  between adjacent terms which are usually written
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These are convergent if  , and divergent if  . If it is convergent, we can find the sum by the formula
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where  is the first term in the series (if the index starts at  or  , then "" is actually the first term  or  , respectively).
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Another common type of series to evaluate is a telescoping series, where the telescoping better describes the partial sums, denoted  . Most of the time, they are presented as a fraction which requires partial fraction decomposition.
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This can be accomplished fairly quickly via a shortcut when the factors in the denominator are linear and share the same coefficient on  .
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Example. Suppose we wish to decompose the fraction  . First, consider the difference
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| If we combine this to a common denominator, we find
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- 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)}}.}
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| 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
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Notice the pattern: for any fraction of the form where we have
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- 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).}}
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| In this manner, we can quickly find that
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- 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)}}
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| As per the so-called telescoping, consider the series defined by
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| Using the technique above, we can rewrite the series as
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- 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=1}^{\infty }{\frac {1}{2}}\left({\frac {1}{n-1}}-{\frac {1}{n+1}}\right).}
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| This means that
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- 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}}}
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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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle n-1}
and  in the denominator. If we then take the limit, we find
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- 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}}.}
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