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Assignment Problems Notice
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If you are looking for some problems with solutions you can find some by clicking on the "Practice Problems" link above.
If you are looking for some problems with solutions you can find some by clicking on the "Practice Problems" link above.
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 \( \displaystyle \sum\limits_{n = 5}^\infty {\frac{{n + 1}}{{{n^3}}}} \).