Saturday, August 10, 2013
Series convergence tests
Partial sum
T∑n=1an is called the Tth partial sum of the series ∞∑n=1an. A series converges if the sequence of partial sums converges.nth term test
If lim, then the series \displaystyle \sum_{n=0}^\infty a_n diverges. Can be used to test for divergence (but not convergence). ie The converse is not always true, but the contra-positive is.Integral test
\begin{aligned} \sum_{n=m}^\infty f(n) \hskip1em \text{converges} \hskip2em \Leftrightarrow \hskip2em \int_{m}^\infty f(x) \, dx \hskip1em \text{converges} \\ \end{aligned} |
p-series test
\begin{aligned} \sum_{n=1}^\infty \frac{1}{n^p} \hskip1em \text{converges if and only if } p > 1 \\ \end{aligned} |
Comparison test
Let a_n and b_n be positive sequences such that b_n > a_n for all n. It follows that if \sum b_n converges, then so does \sum a_n, and if \sum a_n diverges, then so does \sum b_n.
Limit test
Let a_n and b_n be positive series. If \displaystyle \lim_{n \rightarrow \infty} \frac{a_n}{b_n} = L and 0 < L < \infty, then the series \sum a_n and \sum b_n either both converge or both diverge.
Root test
Given a series \sum a_n with all a_n > 0, let \displaystyle \rho = \lim_{n \rightarrow \infty} \sqrt[n]{a_n} . If \rho < 1, then \sum a_n converges. If \rho > 1, then \sum a_n diverges. If \rho = 1, then the test is inconclusive. Based on the idea of "eventual geometric series".
Ratio test
Given a series \sum a_n with all a_n > 0, let \displaystyle \rho = \lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} . If \rho < 1, then \sum a_n converges. If \rho > 1, then \sum a_n diverges. If \rho = 1, then the test is inconclusive. Work best when exponential or factorial factor is involved - not so with ratio of polynomials.
Summary of methods for a positive series
Given a series \sum a_n, where a_n > 0 for all n:
- Do the terms go to 0 ? If \displaystyle \lim_{n \rightarrow \infty} a_n \ne 0 , then by the nth term test, the series diverges. (If \displaystyle \lim_{n \rightarrow \infty} a_n = 0 , then the test is inconclusive).
- Does a_n involve exponential functions like c^n where c is constant ? Does a_n involve factorial ? Then the ratio test should be used.
- Is a_n of the form (b_n)^nfor some sequence b_n ? Then use the root test.
- Does ignoring lower order terms make a_n look like a familiar series (e.g. p-series or geometric series) ? Then use the comparison test or the limit test.
- Does the sequence a_n look easy to integrate ? Then use the integral test.
Alternating Series test
An alternating series \sum (-1)^n b_n with decreasing terms converges if and only if \displaystyle \lim_{n \rightarrow \infty} b_n = 0 . (Note when to examine the asymptotic value of the nth term vs. examining the p-value of an integral.)
Conditional vs Absolute Convergence
\displaystyle \sum a_n converges absolutely if \displaystyle \sum \lvert a_n \rvert converges. \displaystyle \sum a_n converges conditionally if \displaystyle \sum \lvert a_n \rvert diverges but \displaystyle \sum a_n converges. In other words, a series is conditionally convergent if it is convergent but not absolutely convergent.
Absolute Convergence test
If a series converges absolutely, then it always converges. (But this is not necessarily true if it converges conditionally.)
Interesting notes
- \displaystyle \lim_{n \rightarrow \infty} \sqrt[n]{n} = 1
\begin{aligned} y &= \lim_{n \rightarrow \infty} \sqrt[n]{n} = \lim_{n \rightarrow \infty} n^\frac{1}{n} \\ ln(y) &= \lim_{n \rightarrow \infty} \frac{1}{n} ln(n) = 0 \\ \therefore y &= 1 \\ \end{aligned} - \displaystyle \lim_{n \rightarrow \infty} (1 + \frac{1}{n})^n = e = \sum_{n=0}^\infty \frac{1}{n!}
- \displaystyle \lim_{n \rightarrow \infty} (1 + \frac{a}{n})^n = e^a = \sum_{n=0}^\infty \frac{1}{n!} a^n
- \displaystyle 0 < ln(2) < 1