Sean Fitzpatrick |
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University of Lethbridge |
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Decide if the series converges:
\(\di \sum_{n=0}^\infty \frac{5n^7-3n^4+19n^3+2565}{11n^{12}+32n^9+55n^7+2020n^4+3}\)
\(\di \sum_{n=1}^\infty \frac{n^2}{2^n}\)
\(\di \sum_{n=1}^\infty \frac{n!}{n^n}\)
Decide if the series converges:
\(\di \sum_{n=0}^\infty \frac{3^n}{2^n}\)
\(\di \sum_{n=1}^\infty \frac{(n!)^2}{(2n)!}\)
Let \(\{a_n\}\) be a sequence with positive terms, and consider \(\lim_{n\to \infty}\sqrt[n]{a_n}\text{.}\)
If this limit is less than 1, then \(\sum a_n\) converges. If the limit is greater than 1, the series diverges.
Determine the convergence of
\(\sum_{n=1}^\infty \left(\frac{3n+5}{5n+3}\right)^n\)
\(\sum_{n=1}^\infty \frac{1}{n^n}\)
An alternating series is a series of the form \(\di \sum_{n=1}^\infty (-1)^na_n\text{.}\)
If \(\{a_n\}\) is a positive, decreasing sequence with \(\lim\limits_{n\to\infty}a_n=0\text{,}\) then \(\di \sum_{n=1}^\infty (-1)^na_n\) converges.
Examples: \(\di \sum_{n=1}^\infty \frac{(-1)^n}{n}\text{,}\) \(\di\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}}\)
Most of our other convergence tests apply to series with positive terms. To apply the ratio test, etc. to an alternating series, take absolute values.
A series \(\sum a_n\) converges absolutely if \(\sum \abs{a_n}\) converges.
If \(\sum a_n\) converges but \(\sum \abs{a_n}\) does not, we say that the series converges conditionally.
Any absolutely convergent series is convergent.
In an alternating series, the partial sums jump back and forth across the limit.
Each time the jumps get smaller (why?)
Result: let \(L = \lim\limits_{N\to\infty} s_N = \sum_{n=1}^\infty (-1)^na_n\text{.}\) Then
Example: \(\di \sum_{n=1}^\infty \frac{(-1)^n}{n^4} = \frac{\pi^4}{720}\text{.}\) How many terms do we need to take in the series to get an answer accurate to four decimal places?
A power series is a function of the form \(f(x) = \sum_{n=0}^\infty a_n x^n\text{.}\)
The domain for such a function is the set of all \(x\) for which the series converges.
Can also consider power series centred at some number \(c\text{:}\)
\(\di \sum_{n=0}^\infty 2^nx^n\)
\(\di\sum_{n=0}^\infty\frac{(-1)^n}{5^n}(x-3)^n\)
\(\di \sum_{n=0}^\infty \frac{3^nx^n}{n^2}\)
Note that the first two series are geometric. What's the sum?
Applying the ratio test to \(\di\sum_{n=0}^\infty a_n(x-c)^n\) produces the limit
We need this limit to be less than 1.
So we need \(\di \abs{x-c}\lt 1/L\text{,}\) where \(\di L = \lim_{n\to\infty}\abs{\frac{a_{n+1}}{a_n}}\text{.}\)
Define \(R=1/L\) as the radius of convergence. (\(R=0\) if \(L=\infty\) and \(R=\infty\) if \(L=0\text{.}\))
Determine radius and interval of convergence:
\(\di \sum_{n=0}^\infty (-2)^nx^n\)
\(\di \sum_{n=0}^\infty \frac{1}{n}(x-2)^n\)
\(\di \sum_{n=0}^\infty \frac{(-1)^n}{n^3}x^n\)
\(\di \sum_{n=0}^\infty \frac{(2x)^n}{n!}\)
Did we get to this before time ran out? Let's do some examples on the board!