Paul's Online Notes
Home / Calculus II / Series & Sequences / Series - The Basics
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

### 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 $$\displaystyle \sum\limits_{n = 2}^\infty {\frac{1}{{{n^3} + 1}}}$$.

Show All Steps Hide All Steps

Start Solution

First 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 2

Now, 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,

$1.6865 = \frac{3}{2} + \sum\limits_{n = 2}^\infty {\frac{1}{{{n^3} + 1}}} \hspace{0.25in}\hspace{0.25in} \Rightarrow \hspace{0.25in}\hspace{0.25in}\sum\limits_{n = 2}^\infty {\frac{1}{{{n^3} + 1}}} = 1.6865 - \frac{3}{2} = \require{bbox} \bbox[2pt,border:1px solid black]{{0.1865}}$