Now, we need to be careful here. There is no way to actually identify the series as a telescoping series at this point. We are only hoping that it is a telescoping series.

Therefore, the first real step here is to perform partial fractions on the series term to see what we get. Here is the partial fraction work for the series term.

\[\frac{{10}}{{{n^2} - 4n + 3}} = \frac{{10}}{{\left( {n - 1} \right)\left( {n - 3} \right)}} = \frac{A}{{n - 1}} + \frac{B}{{n - 3}}\hspace{0.25in}\hspace{0.25in} \to \,\,\,\,\,\,\,\,\,\,\,10 = A\left( {n - 3} \right) + B\left( {n - 1} \right)\] \[\begin{array}{l}{n = 1}\hspace{0.5in}{10 = - 2A}\\{n = 3}\hspace{0.5in}{10 = 2B}\end{array}\hspace{0.25in} \to \hspace{0.25in}\hspace{0.25in}\begin{array}{l}{A = - 5}\\{B = 5}\end{array}\]

The series term in partial fraction form is then,

\[\frac{{10}}{{{n^2} - 4n + 3}} = \frac{5}{{n - 3}} - \frac{5}{{n - 1}}\] Show Step 3

The partial sums for this series are then,

\[{s_n} = \sum\limits_{i = 4}^n {\left[ {\frac{5}{{i - 3}} - \frac{5}{{i - 1}}} \right]} \] Show Step 4

Expanding the partial sums from the previous step give,

\[\begin{align*}{s_n} & = \sum\limits_{i = 4}^n {\left[ {\frac{5}{{i - 3}} - \frac{5}{{i - 1}}} \right]} \\ & = \left[ {\frac{5}{1} - \require{cancel} \cancel{{\frac{5}{3}}}} \right] + \left[ {\frac{5}{2} - \require{cancel} \cancel{{\frac{5}{4}}}} \right] + \left[ {\require{cancel} \cancel{{\frac{5}{3}}} - \require{cancel} \bcancel{{\frac{5}{5}}}} \right] + \left[ {\require{cancel} \cancel{{\frac{5}{4}}} - \require{cancel} \bcancel{{\frac{5}{6}}}} \right] + \left[ {\require{cancel} \bcancel{{\frac{5}{5}}} - \require{cancel} \cancel{{\frac{5}{7}}}} \right] + \cdots \\ & \hspace{1.0in} + \left[ {\require{cancel} \cancel{{\frac{5}{{n - 7}}}} - \require{cancel} \bcancel{{\frac{5}{{n - 5}}}}} \right] + \left[ {\require{cancel} \bcancel{{\frac{5}{{n - 6}}}} - \require{cancel} \cancel{{\frac{5}{{n - 4}}}}} \right] + \left[ {\require{cancel} \bcancel{{\frac{5}{{n - 5}}}} - \require{cancel} \cancel{{\frac{5}{{n - 3}}}}} \right] + \\ & \hspace{1.5in}\left[ {\require{cancel} \cancel{{\frac{5}{{n - 4}}}} - \frac{5}{{n - 2}}} \right] + \left[ {\require{cancel} \cancel{{\frac{5}{{n - 3}}}} - \frac{5}{{n - 1}}} \right]\\ & = 5 + \frac{5}{2} - \frac{5}{{n - 2}} - \frac{5}{{n - 1}}\end{align*}\]

It is important when doing this expanding to expand out from both the initial and final values of \(i\) and to expand out until all the parts of a series term cancel. Once that has been done it is safe to assume that the cancelling will continue until we get near the end of the expansion.

Also, as seen above these can be quite messy to expand out until everything starts to cancel out so don’t get too excited about it when it does get messy like this. It just happens sometimes and we have to be careful with all the expansion.

Note that at this point we now know that the series was a telescoping series since we got almost all the “interior” terms to cancel out.

Show Step 5

At this point all we need to do is look at the limit of the partial sums to get,

\[\mathop {\lim }\limits_{n \to \infty } {s_n} = \mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{15}}{2} - \frac{5}{{n - 2}} - \frac{5}{{n - 1}}} \right] = \frac{{15}}{2}\] Show Step 6

The limit of the partial sums exists and is a finite number (i.e. not infinity) and so we can see that the series converges and its value is,

\[\sum\limits_{n = 4}^\infty {\frac{{10}}{{{n^2} - 4n + 3}}} = \frac{{15}}{2}\]