Section 10.3 : Series - Basics
For problems 1 – 3 perform an index shift so that the series starts at \(n = 3\).
- \( \displaystyle \sum\limits_{n = 1}^\infty {\left( {n{2^n} - {3^{1 - n}}} \right)} \) Solution
- \( \displaystyle \sum\limits_{n = 7}^\infty {\frac{{4 - n}}{{{n^2} + 1}}} \) Solution
- \( \displaystyle \sum\limits_{n = 2}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 3}}\left( {n + 2} \right)}}{{{5^{1 + 2n}}}}} \) Solution
- Strip out the first 3 terms from the series \( \displaystyle \sum\limits_{n = 1}^\infty {\frac{{{2^{ - n}}}}{{{n^2} + 1}}} \). Solution
- Given that \( \displaystyle \sum\limits_{n = 0}^\infty {\frac{1}{{{n^3} + 1}}} = 1.6865\) determine the value of \( \displaystyle \sum\limits_{n = 2}^\infty {\frac{1}{{{n^3} + 1}}} \). Solution