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### Section 10.3 : Series - Basics

4. Strip out the first 3 terms from the series $$\displaystyle \sum\limits_{n = 1}^\infty {\frac{{{2^{ - n}}}}{{{n^2} + 1}}}$$.

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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.