Class 20
\[\sum_{i=0}^\infty ar^i\]
Converges to \(\displaystyle\frac{a}{1-r}\) if \(\left\lvert r\right\rvert < 1\).
Diverges if \(\left\lvert r\right\rvert \ge 1\)
\[\sum_{i=k}^\infty \left(b_i - b_{i+1}\right)\]
\(s_n = b_k - b_{n+1}\)
Converges to \(b_k - \displaystyle\lim_{n\to\infty} b_n\) if this limit exists.
A discrete version of the FTC!
\(\displaystyle \sum_{i = 1}^\infty \frac{(i+1)\sin(i) - i\sin(i+1)}{i(i+1)}\)
\[\sum_{i = 1}^\infty \frac{1}{i}\]
divergent
\(\displaystyle s_{2^n} > \frac{n}{2}\)
If the \[\lim_{n\to\infty} a_n\] is not 0, then the series \[\sum_{i=k}^\infty a_i\] diverges.
Let \(N > k\) are two integers. Then \(\displaystyle \sum_{i=k}^\infty a_i\) converges if and only if \(\displaystyle\sum_{i=N}^\infty a_i\) converges.
Proof: Let \(s_n\), \(n = k, k+1, \dots\) be the sequence of partial sums of the series \(\displaystyle \sum_{i=k}^\infty a_i\), and let \(r_n\), \(n=N, N+1, \dots\) be the sequence of partial sums of the series \(\displaystyle \sum_{i=N}^\infty a_i\).
Then \(s_n = r_n + s_{N-1}\) for every \(n \ge N\).
If \(a_i \ge 0\) for every \(i = k, k+1, \dots\), we can visualize the series as an area:
If \(0 \le a_n \le b_n\) for each \(n \ge k\) and if \(\displaystyle\sum_{i=k}^\infty b_i\) converges, then \(\displaystyle\sum_{i=k}^\infty a_i\) also converges.
If \(a_n \ge b_n \ge 0\) for each \(n \ge k\) and if \(\displaystyle\sum_{i=k}^\infty b_i\) diverges, then \(\displaystyle\sum_{i=k}^\infty a_i\) also diverges.
\(\displaystyle\sum_{i=0}^\infty \frac{1}{3^i + 5}\)
\(\displaystyle\sum_{i=2}^\infty \frac{1}{i-1}\)
Suppose \(a_n = f(n)\) for \(n = k, k+1, \dots\), where \(f\) is a decreasing and continuous function.
Suppose \(a_n = f(n)\) for \(n = k, k+1, \dots\), where \(f\) is a decreasing and continuous function.
If \(\displaystyle\int_{k-1}^\infty f(x)\;dx\) converges, then so does \(\displaystyle \sum_{i=k}^\infty a_i\).
If \(\displaystyle\int_{k}^\infty f(x)\;dx\) diverges, then so does \(\displaystyle \sum_{i=k}^\infty a_i\).
\(\displaystyle\sum_{i=0}^\infty \frac{1}{i^2 + 1}\)
\(\displaystyle\sum_{i=2}^\infty \frac{1}{i\ln i}\)
\(\displaystyle\sum_{i=1}^\infty \frac{1}{i^p}\)
\(\displaystyle\sum_{i=1}^\infty \frac{1}{i^2 + 1}\)
\(\displaystyle\sum_{i=2}^\infty \frac{1}{i^2 - 1}\)
\(\displaystyle\sum_{i=2}^\infty \frac{1}{i - 1}\)
\(\displaystyle\sum_{i=1}^\infty \frac{1}{i + 1}\)
Suppose \(a_n \ge 0\) and \(b_n \ge 0\) for all \(n \ge k\).
\(\displaystyle\sum_{i=2}^\infty \frac{1}{i^2 - 1}\)
\(\displaystyle\sum_{i=1}^\infty \frac{1}{i + 1}\)
\(\displaystyle\sum_{i=1}^\infty \frac{i}{i^2 + 1}\)