Partial sum

\(\displaystyle \sum_{n=1}^T a_n\) is called the \(T\)th partial sum of the series \(\displaystyle \sum_{n=1}^\infty a_n\). A series converges if the sequence of partial sums converges.

\(n\)th term test

If \(\displaystyle \lim_{n \rightarrow \infty} a_n \ne 0\), 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

\[ \begin{aligned} \sum_{n=m}^\infty f(n) \hskip1em \text{converges} \hskip2em \Leftrightarrow \hskip2em \int_{m}^\infty f(x) \, dx \hskip1em \text{converges} \\ \end{aligned} \]

Condition: \(f(x)\) is a positive decreasing function.

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 \(n\):

1. 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).
2. 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.
3. Is \(a_n\) of the form \((b_n)^n\)for some sequence \(b_n\) ? Then use the root test.
4. 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.
5. 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 \(n\)th 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

• \(\displaystyle \lim_{n \rightarrow \infty} \sqrt[n]{n} = 1\)

  \[ \begin{aligned} y &= \lim_{n \rightarrow \infty} \sqrt[n]{n} = \lim_{n \rightarrow \infty} n^\frac{1}{n} \\ ln(y) &= \lim_{n \rightarrow \infty} \frac{1}{n} ln(n) = 0 \\ \therefore y &= 1 \\ \end{aligned} \]

• \(\displaystyle \lim_{n \rightarrow \infty} (1 + \frac{1}{n})^n = e = \sum_{n=0}^\infty \frac{1}{n!} \)
• \(\displaystyle \lim_{n \rightarrow \infty} (1 + \frac{a}{n})^n = e^a = \sum_{n=0}^\infty \frac{1}{n!} a^n \)
• \(\displaystyle 0 < ln(2) < 1 \)