Given the sequence \(u_1, u_2, \ldots, u_n, \ldots\), the sum \(u_1 + u_2 + \cdots + u_n + \cdots\) is called the infinite series or series, denoted by \(\displaystyle \sum_{i=1}^{\infty} u_n\), where the \(u_n\) is called the general term of the series.
We define the sum of infinite series by its partial sum sequence:
If the partial sum sequence \(\{s_n\}, s_n := \displaystyle \sum_{i=1}^{n}u_i\) of series \(\displaystyle \sum_{i=1}^{\infty} u_n\) has limit \(s = \displaystyle \lim_{n\rightarrow\infty} s_n\), then we say the series converges, and the limit \(s\) is defined as the sum of the series, otherwise we say the series diverges.
The difference between the limit and the partial sum is called the remainder terms \(r_n\): \[ r_n = s - s_n = \sum_{i=n+1}^{\infty} u_i \] If we use partial sum \(s_n\) to estimate the sum of series \(s\), then the error will be absolute value of remainder \(|r_n|\).
Linearity. If series \(\displaystyle \sum_{i=1}^{\infty} u_n, \sum_{i=1}^{\infty} v_n\) converges to \(s, \sigma\) respectively, then the series \(\displaystyle \sum_{i=1}^{\infty} (k_1u_n \pm k_2v_n)\) converges to \(k_1s + k_2\sigma\). This can be inferred by the linearity of the limit of sequence immediately.
Adding or removing finite term will not change the convergene of series.
Adding arbitrary parentheses to the series will not change the convergence of series.
Corollary. If the series with parentheses added is divergent, then the primitive series is also divergent.
Example. Prove the harmonic series \(\displaystyle \sum_{i=1}^{\infty} \frac{1}{i}\) is divergent.
Proof. \[ \begin{align} \sum_{i=1}^{\infty} \frac{1}{i} &= 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} \\ &= 1 + \frac{1}{2} + \left( \frac{1}{3} + \frac{1}{4} \right) + \left( \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} \right) + \cdots\\ &\ge 1 + \frac{1}{2} + 2\times\frac{1}{4} + 4\times \frac{1}{8} + \cdots \\ &\rightarrow \infty \end{align} \] hence the harmonic series is also divergent.
If the series \(\displaystyle \sum_{i=1}^{\infty} u_n\) is convergent, then its general term \(u_n\) approaches zero, that is, \(\displaystyle \lim_{n\rightarrow\infty} u_n = 0\).
Proof. Denotes the partial sum of convergent sereis \(\displaystyle \sum_{i=1}^{\infty} u_n\) as \(s_n\), we haves \[ \displaystyle \lim_{n\rightarrow\infty} u_n = \lim_{n\rightarrow\infty} (s_n -s_{n-1}) = \lim_{n\rightarrow\infty} s_n - \lim_{n\rightarrow\infty} s_{n-1} = s - s = 0 \]
Corollary. If the general term of series not approaches zero, then the series must be divergent.
Remark. The general term approaches zero is not the sufficient condition of convergent, the harmonic series \(\displaystyle \sum_{i=1}^{\infty} \frac{1}{i}\) is a typical example.
The series \(\displaystyle \sum_{i=1}^{\infty} u_n\) converge, if any only if, for any given positive number \(\varepsilon\), there always exists an positive number \(N\) that gurantees \(|u_{n+1} + u_{n+2} + \cdots + u_{n+p}| < \varepsilon\) for any \(p\in \mathbb{Z}^+\) when \(n>N\).
As it named, positive series is the series with all its item are non-negative. In this case, the partial sum sequence \(\{s_n\}\) will be an increasing sequence.
Since \(\{s_n\}\) is monotone, based on the property of limit of sequence, the limit of \(\{s_n\}\) exists if and only if the \(\{s_n\}\) is bounded. That is, The series is convergent, if and only if its partial sum sequence \(\{s_n\}\) is bounded.
The Direct comparison test can be introduced based on the bound theorem. For two positive item series, if there exsits \(N\in \mathbb{Z}^+\) which gurantees for any \(n>N\), \(u_n\le v_n\), then
if \(\displaystyle \sum_{i=1}^{\infty} v_n\) is convergent, then \(\displaystyle \sum_{i=1}^{\infty} u_n\) is convergent.
if \(\displaystyle \sum_{i=1}^{\infty} u_n\) is divergent, then \(\displaystyle \sum_{i=1}^{\infty} v_n\) is divergent.
and this can be extended to an limit form, named as limit comparison test:
For two positive item series \(\displaystyle \sum_{i=1}^{\infty} u_n, \sum_{i=1}^{\infty} v_n\), compute the limit of the ratio \(L = \displaystyle \lim_{n\rightarrow\infty} \dfrac{u_n}{v_n}\), then
Remark. This indicates that for \(L\in(0, +\infty)\), the two series share the same convergence.
Example. Discuss whether the series \(\displaystyle \sum_{i=1}^{\infty} \sin \dfrac{1}{n}\) converge.
\[ \lim_{n\rightarrow \infty} \frac{\sin \dfrac{1}{n}}{\dfrac{1}{n}} = 1 \]
and we’ve already known the harmonic series is divergent, hence the series \(\displaystyle \sum_{i=1}^{\infty} \sin \dfrac{1}{n}\) is also divergent.
Example. Discuss whether the series \(\displaystyle \sum_{i=1}^{\infty} \ln\left(1+\frac{1}{n^2}\right)\) converge.
\[ \lim_{n\rightarrow \infty} \frac{\ln\left(1+\dfrac{1}{n^2}\right)}{\dfrac{1}{n^2}} = 1 \]
and we’ve already known the series \(\displaystyle \sum_{i=1}^{\infty} \frac{1}{n^2}\) is convergent, hence the given series is also convergent.
For the positive term series \(\displaystyle \sum_{i=1}^{\infty} u_n\), compute \(\rho = \displaystyle \lim_{n\rightarrow\infty} \dfrac{u_{n+1}}{u_n}\).
For the positive term series \(\displaystyle \sum_{i=1}^{\infty} u_n\), compute \(\rho = \displaystyle \lim_{n\rightarrow\infty} \sqrt[n]{n}\).
Example. determine whether the following series converge or diverge.
- \(\displaystyle \sum \frac{e^nn!}{n^n}\)
\[ \frac{u_{n+1}}{u_n} = \frac{e}{\left(1+\dfrac{1}{n}\right)^n} > 1 \]
that is, \(u_n>u_{n-1} > \cdots > u_1 = e\), hence \(\displaystyle \lim_{n\rightarrow\infty} u_n\neq 0\).
- \(\displaystyle \sum \frac{1}{n^3+3n^2+2n}\)
\[ \sum \frac{1}{n^3+3n^2+2n} = \frac{1}{n(n+1)(n+2)} \]
- \(\displaystyle \sum \frac{2n-1}{2^n}\)
\[ S_n = 3 - \frac{1}{2^{n-2}} - \frac{2n-1}{2^n} \]
hence \(\displaystyle \lim_{n\rightarrow \infty} u_n =3\).
Theorem. The \(\displaystyle \sum_{i=1}^{\infty} u_n\) converge if any only if \(S_n\) is bounded.
Example. Discuss whether the series \(1 + \dfrac{1}{2^p} + \dfrac{1}{3^p} + \cdots + \dfrac{1}{n^p} + \cdots\) converge
The alternating series is the series in the form \(\displaystyle \sum_{n=1}^{\infty} (-1)^{n-1}u_n\) or \(\displaystyle \sum_{n=1}^{\infty} (-1)^{n}u_n\). That is, the series with alternating positive number and negative number.
For the alternating series \(\displaystyle \sum_{n=1}^{\infty} (-1)^{n-1}u_n\), if \(\displaystyle u_ n\ge u_{n+1}, \lim_{n\rightarrow\infty} u_n=0\), then the series converges, and the absolute value of remainder \(|r_n|\le u_{n+1}\).
Proof.
Example. The alternating harmonic series \(\displaystyle \sum_{n=1}^{\infty} (-1)^{n-1}\frac{1}{n}\) converges, since \(\dfrac{1}{n}\ge \dfrac{1}{n+1}\) and \(\displaystyle \lim_{n\rightarrow\infty} \dfrac{1}{n} = 0\).
For the series \(\displaystyle \sum_{i=1}^{\infty} u_n\), if the positive term series \(\displaystyle \sum_{i=1}^{\infty} |u_n|\) converges, then we say the series \(\displaystyle \sum_{i=1}^{\infty} u_n\) is converges absolutely. If the series \(\displaystyle \sum_{i=1}^{\infty} u_n\) converges but the positive term series \(\displaystyle \sum_{i=1}^{\infty} |u_n|\) disverges, then we say the series \(\displaystyle \sum_{i=1}^{\infty} u_n\)converges conditionally. For example, the series \(\displaystyle \sum_{n=1}^{\infty} (-1)^{n-1}\frac{1}{n^2}\) converges absoluately, while the series \(\displaystyle \sum_{n=1}^{\infty} (-1)^{n-1}\frac{1}{n}\) converges conditionally.