Sean Fitzpatrick |
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University of Lethbridge |
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Decide if the series converges:
\(\di \sum_{n=0}^\infty e^{-n}\)
\(\di \sum_{n=1}^\infty \frac{1}{\sqrt{n}}\)
\(\di \sum_{n=1}^\infty \frac{1}{n^2}\)
Sometimes, terms in a series cancel (if we're lucky).
Determine the converge of:
\(\di \sum_{n=1}^\infty\frac{1}{n^2+n}\)
\(\di \sum_{n=4}^\infty \frac{1}{n^2-3n}\)
Let \(\sum a_n\) and \(\sum b_n\) be convergent series. Then:
\(\di \sum(a_n\pm b_n) = \sum a_n \pm \sum b_n\)
For any constant \(c\text{,}\) \(\di\sum ca_n = c\sum a_n\)
Evaluate the sums
\(\di \sum_{n=2}^\infty\left(\frac{1}{4^n}-\frac{3}{n^2}\right)\)
\(\di\frac18+\frac{1}{16}+\frac{1}{32}+\cdots\)
\(\di \frac19-\frac{1}{16}+\frac{1}{25}-\frac{1}{36}-\cdots\)
The integral test lets us use what we already learned from improper integrals:
Let \(f\) be a positive, continuous, decreasing function on \([1,\infty)\text{,}\) and let \(\{a_n\} = \{f(n)\}\text{.}\) Then \(\di\sum_{n=1}^\infty a_n\) converges if and only if \(\di\int_1^\infty f(x)\,dx\) converges.
Example: \(\di \sum_{n=1}^\infty \frac{1}{n\ln(n)}\)
A \(p\)-series is a series of the form
A \(p\)-series:
converges, if \(p>1\)
diverges, if \(p\leq 1\)
A famous \(p\)-series is \(\di \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}\text{.}\)
If the series \(\di\sum_{n=1}^\infty a_n\) converges, then \(\di\lim_{n\to\infty}a_n=0\text{.}\)
The series \(\di\sum_{n=1}^\infty \frac1n\) is called the harmonic series.
Despite the fact that \(\frac1n\to 0\) as \(n\to\infty\text{,}\) the harmonic series diverges.
However, the alternating harmonic series \(\di \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}\) converges (to \(\ln(2)\))!
Let \(\{a_n\}=\{f(n)\}\) be a sequence with positive terms for which the integral test applies, and \(\sum a_n\) converges. Let
Approximate \(\sum_{n=1}^\infty\frac{1}{n^4}\) using the first 5 terms.
Works the same as comparison for improper integrals.
Let \(\{a_n\},\{b_n\}\) be sequences with positive terms, such that \(a_n\leq b_n\) for all \(n\text{.}\)
If \(\sum a_n\) diverges, then \(\sum b_n\) diverges.
If \(\sum b_n\) converges, then \(\sum a_n\) converges.
\(\di\sum_{n=1}^\infty \frac{1}{n^2+n}\)
\(\di \sum_{n=1}^\infty \frac{1}{\sqrt{2n^2-1}}\)
\(\di\sum_{n=2}^\infty \frac{1}{n^3-1}\)
Let \(\{a_n\},\{b_n\}\) be sequences with positive terms, such that \(a_n\leq b_n\) for all \(n\text{.}\) If
Determine the convergence of
\(\sum_{n=1}^\infty\frac{3n^3-7n^2+5}{9n^7+8n^4+2565}\)
\(\sum_{n=1}^\infty \frac{1}{n+\ln(n)}\)
Let \(\{a_n\}\) be a sequence with positive terms. Consider
Determine the convergence of
\(\di\sum_{n=1}^\infty\frac{2^n}{n!}\)
\(\di\sum_{n=1}^\infty \frac{1}{n^2+4n}\)
\(\di\sum_{n=1}^\infty \frac{3n!}{(3n)!}\)
\(\di \frac{2\sqrt{2}}{9801}\sum_{n=1}^\infty \frac{(4n)!(1103+26390n)}{(n!)^4 396^{4n}}\)
Let \(\{a_n\}\) be a sequence with positive terms, and consider
Determine the convergence of
\(\sum_{n=1}^\infty \left(\frac{3n+5}{5n+3}\right)\)
\(\sum_{n=1}^\infty \frac{1}{n^n}\)