Remember that when we say we are going to “strip out” terms from a series we aren’t really getting rid of them. All we are doing is writing the first few terms of the series as a summation in front of the series.

So, for this series stripping out the first three terms gives,

\[\begin{align*}\sum\limits_{n = 1}^\infty {\frac{{{2^{ - n}}}}{{{n^2} + 1}}} & = \frac{{{2^{ - 1}}}}{{{1^2} + 1}} + \frac{{{2^{ - 2}}}}{{{2^2} + 1}} + \frac{{{2^{ - 3}}}}{{{3^2} + 1}} + \sum\limits_{n = 4}^\infty {\frac{{{2^{ - n}}}}{{{n^2} + 1}}} \\ & = \frac{1}{4} + \frac{1}{{20}} + \frac{1}{{80}} + \sum\limits_{n = 4}^\infty {\frac{{{2^{ - n}}}}{{{n^2} + 1}}} \\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{5}{{16}} + \sum\limits_{n = 4}^\infty {\frac{{{2^{ - n}}}}{{{n^2} + 1}}} }}\end{align*}\]

This first step isn’t really all that necessary but was included here to make it clear that we were plugging in \(n = 1\), \(n = 2\) and \(n = 3\) (i.e. the first three values of \(n\)) into the general series term. Also, don’t forget to change the starting value of \(n\) to reflect the fact that we’ve “stripped out” the first three values of \(n\) or terms.