Class 19
Definition: Given an infinite sequence \(a = (a_n)_{n=k,\dots,\infty}\), we define the infinite series associated with \(a\) the formal sum \[a_k + a_{k+1} + a_{k+2} + \cdots = \sum_{i = k}^\infty a_i\] In other words, we place a \(+\) sign between each two consecutive terms of the sequence.
Question: Can this be interpreted as an actual sum, in other words, can we actually add all infinitely many terms of the sequence?
\(\displaystyle a_n = \frac{1}{2^n}\) for \(n = 1, 2, dots\).
Given an infinite series \(\displaystyle\sum_{i = k}^\infty a_i = a_k + a_{k+1} + a_{k+2} + \cdots\), we define a sequence of partial sums of the series, \(s_n\) for \(n = k, k+1, \dots\) : \[s_n = \sum_{i = k}^n a_i = a_k + a_{k+1} + a_{k+2} + \cdots + a_n\]
If the sequence \(s_n\) converges, we say that the series \(\displaystyle \sum_{i = k}^\infty a_i\) is convergent, and we define \[\displaystyle \sum_{i = k}^\infty a_i = \lim_{n\to\infty} s_n\]
Otherwise, we say that the series diverges or is divergent.
\(\displaystyle \sum_{i=1}^\infty\frac{1}{2^i}\)
The following series are all the same:
If \(\displaystyle\lim_{n\to\infty} a_n \neq 0\) then \(\displaystyle\sum_{i=k}^\infty a_i\) diverges.
This is known as the \(n\)-th term test for divergence.
\(\displaystyle \sum_{i = 1}^\infty \frac{1}{n}\)
Trick: Look at the partial sums \(s_n\) where \(n\) is a power of 2:
\(\displaystyle \sum_{i=1}^\infty \frac{1}{i^2 + i}\)
\(\displaystyle \sum_{i=0}^\infty a\cdot r^i\)
Suppose \(\displaystyle\sum_{i=k}^\infty a_i\) and \(\displaystyle\sum_{i=k}^\infty b_i\) are both convergent.
Then:
\(\displaystyle \sum_{i = 1}^\infty \frac{1}{2^i} + \frac{7}{3^{i-1}}\)
Suppose \(a_n \ge 0\) for every \(n \ge k\). Then the sequence of partial sums of the series \(\displaystyle\sum_{i=k}^\infty a_i\) is increasing.
Then the series \(\displaystyle\sum_{i=k}^\infty a_i\) converges if and only if the sequence of its partial sums is bounded.
\(\displaystyle\sum_{i=0}^\infty \frac{1}{2^i + 1}\)