Section 10.3 : Series - Basics
5. Given that \( \displaystyle \sum\limits_{n = 0}^\infty {\frac{1}{{{n^3} + 1}}} = 1.6865\) determine the value of \( \displaystyle \sum\limits_{n = 2}^\infty {\frac{1}{{{n^3} + 1}}} \).
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Start SolutionFirst notice that if we strip out the first two terms from the series that starts at \(n = 0\) the result will involve a series that starts at \(n = 2\).
Doing this gives,
\[\sum\limits_{n = 0}^\infty {\frac{1}{{{n^3} + 1}}} = \frac{1}{{{0^3} + 1}} + \frac{1}{{{1^3} + 1}} + \sum\limits_{n = 2}^\infty {\frac{1}{{{n^3} + 1}}} = \frac{3}{2} + \sum\limits_{n = 2}^\infty {\frac{1}{{{n^3} + 1}}} \] Show Step 2Now, for this situation we are given the value of the series that starts at \(n = 0\) and are asked to determine the value of the series that starts at \(n = 2\). To do this all we need to do is plug in the known value of the series that starts at \(n = 0\) into the “equation” above and “solve” for the value of the series that starts at \(n = 2\).
This gives,
\[1.6865 = \frac{3}{2} + \sum\limits_{n = 2}^\infty {\frac{1}{{{n^3} + 1}}} \hspace{0.25in}\hspace{0.25in} \Rightarrow \hspace{0.25in}\hspace{0.25in}\sum\limits_{n = 2}^\infty {\frac{1}{{{n^3} + 1}}} = 1.6865 - \frac{3}{2} = \require{bbox} \bbox[2pt,border:1px solid black]{{0.1865}}\]