Section 10.3 : Series - Basics
For problems 1 – 4 perform an index shift so that the series starts at \(n = 4\).
- \( \displaystyle \sum\limits_{n = 8}^\infty {\frac{{2 + n}}{{5 - n}}} \)
- \( \displaystyle \sum\limits_{n = 2}^\infty {\frac{{{3^{n + 2}}}}{{1 - {4^{3n + 1}}}}} \)
- \( \displaystyle \sum\limits_{n = 0}^\infty {{{\left( { - 2} \right)}^{2 - n}}\,{{\bf{e}}^{3n}}} \)
- \( \displaystyle \sum\limits_{n = 5}^\infty {\frac{{{{\left( { - 1} \right)}^{3 + n}}{n^2}}}{{{n^2} - 2n + 1}}} \)
- Strip out the first 4 terms from the series \( \displaystyle \sum\limits_{n = 0}^\infty {{3^n}\,{6^{2 - n}}} \).
- Strip out the first 2 terms from the series \( \displaystyle \sum\limits_{n = 3}^\infty {\frac{4}{{{n^2} + n + 1}}} \).
- Given that \( \displaystyle \sum\limits_{n = 4}^\infty {n\,{4^{ - n}}} = 0.02257\) determine the value of \( \displaystyle \sum\limits_{n = 1}^\infty {n\,{4^{ - n}}} \).
- Given that \( \displaystyle \sum\limits_{n = 3}^\infty {\frac{{n + 1}}{{{n^3}}}} = 0.47199\) determine the value of \( \displaystyle \sum\limits_{n = 5}^\infty {\frac{{n + 1}}{{{n^3}}}} \).