Calculus II - Convergence/Divergence of Series (Practice Problems)

Hide Mobile Notice

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 viewed in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (you should be able to scroll/swipe to see them) and some of the menu items will be cut off due to the narrow screen width.

Section 10.4 : Convergence/Divergence of Series

For problems 1 & 2 compute the first 3 terms in the sequence of partial sums for the given series.

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

For problems 3 & 4 assume that the $n$ ^th term in the sequence of partial sums for the series $\displaystyle \sum\limits_{n = 0}^\infty {{a_n}}$ is given below. Determine if the series $\displaystyle \sum\limits_{n = 0}^\infty {{a_n}}$ is convergent or divergent. If the series is convergent determine the value of the series.

$\displaystyle {s_n} = \frac{{5 + 8{n^2}}}{{2 - 7{n^2}}}$ Solution
$\displaystyle {s_n} = \frac{{{n^2}}}{{5 + 2n}}$ Solution

For problems 5 & 6 show that the series is divergent.

$\displaystyle \sum\limits_{n = 0}^\infty {\frac{{3n\,{{\bf{e}}^n}}}{{{n^2} + 1}}}$ Solution
$\displaystyle \sum\limits_{n = 5}^\infty {\frac{{6 + 8n + 9{n^2}}}{{3 + 2n + {n^2}}}}$ Solution