Sean Fitzpatrick |
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University of Lethbridge |
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Find the limit of the sequence:
\(\di \lim_{n\to\infty}\frac{2^n}{3^n}\)
\(\di \lim_{n\to\infty}(\ln(n+1)-\ln(n))\)
Decide if the following geometric series converge.
\(\displaystyle \sum_{n=0}^\infty \left(\frac35\right)^n\)
\(\displaystyle \sum_{n=0}^\infty \left(\frac53\right)^n\)
Limit properties are pretty much what you expect. Assume \(\{a_n\}\) and \(\{b_n\}\) converge. Then:
\(\di \lim_{n\to\infty}(a_n\pm b_n) = \lim_{n\to \infty}a_n \pm \lim_{n\to\infty}b_n\)
\(\di \lim_{n\to \infty}ca_n = c\lim_{n\to \infty}a_n\)
\(\di \lim_{n\to \infty}(a_nb_n) = \lim_{n\to \infty}a_n\cdot\lim_{n\to\infty}b_n\)
If \(f\) is continuous, \(\lim\limits_{n\to \infty}f(a_n)=f\left(\lim\limits_{n\to\infty}a_n\right)\text{.}\)
A sequence \(\{a_n\}\) is bounded if \(m\lt a_n \lt M\) for some real \(m,M\text{.}\)
A sequence is monotone increasing if \(a_n\lt a_{n+1}\) for all \(n\text{.}\)
A sequence is monotone decreasing if \(a_n\gt a_{n+1}\) for all \(n\text{.}\)
Any convergent sequence is bounded.
Any bounded, monotone sequence converges.
Of the following sequences, which are bounded? Which are montone?
\(\di a_n = \frac{5}{2n+1}\)
\(\di a_n = \frac{2n-1}{3n+5}\)
\(\di a_n = \sin(n\pi/4)\)
\(\di a_n = \frac{n^2-4}{3n}\)
Let \(\{a_n\}\) be defined as follows:
Show that \(\{a_n\}\) converges, and find its limit.
Given a sequence \(\{a_n\}\text{,}\) construct a new sequence \(\{s_n\}\) by
In general, \(s_n = \sum_{k=1}^n a_k\text{.}\)
A series is the limit of a sequence of partial sums. We write
If the limit exists (and is finite) we say the series converges. Otherwise, it diverges.
\(\sum_{n=0}^\infty (-1)^n\)
Geometric series have the form
Question: when does the series converge?
Variation:
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 \(\ln(2)\))!
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
\(\di \sum_{n=2}^\infty\left(\frac{1}{4^n}-\frac{3}{n^2}\right)\)
\(\di\frac18+\frac{1}{16}+\frac{1}{32}+\cdots\)
\(\di \frac19-\frac{1}{16}+\frac{1}{25}-\frac{1}{36}-\cdots\)