Real Series#

Definition: Partial Sum

Let \((a_n)_{n \in \mathcal{D}}\) be a real sequence.

The \(k\)-th partial sum of \((a_n)_{n \in \mathcal{D}}\) is the sum of its first \(k\) numbers:

\[\sum_{\begin{aligned}j &\in \mathcal{D} \\ j &\le k \end{aligned}} a_j\]

Notation

\[s_k\]

Theorem: Re-Indexing

\[\sum_{k = l}^u a_k = \sum_{j = l + r}^{u+r} a_{j - r}\]

Proof

TODO

Given a real sequence \((a_n)_{n \in \mathcal{D}}\), it is apparent that the real-valued function which maps each \(k \in \mathcal{D}\) to the \(k\)-th partial sum of \((a_n)_{n \in \mathcal{D}}\) is itself a real sequence \((s_n)_{n \in \mathcal{D}}\).

Definition: Real Series

A real series is the sequence of partial sums of a real sequence.

Notation

If the sequence is \((a_n)_{n \in \mathcal{D}}\), then we denote the sequence of its partial sums by \(\displaystyle \sum_{n \in \mathcal{D}} a_n\).

When \(\mathcal{D} = \mathbb{N}_0\) or \(\mathcal{D} = \mathbb{N}\), we write \(\displaystyle \sum_{n = 0}^{\infty} a_n\) or \(\displaystyle \sum_{n = 1}^{\infty} a_n\), respectively.

In the case, when \(\mathcal{D} = \mathbb{N} \setminus \{0, 1, \dotsc, p - 1\}\) for some integer \(p\), we write \(\displaystyle \sum_{n = p}^{\infty} a_n\).

Convergence#

Since a real series \(\displaystyle \sum_{n \in \mathcal{D}} a_n\) is just a real sequence, the definitions and terminology for convergence of series is the same as those used for convergence of sequences. What is different, however, is the notation used.

Notation: Convergence of Real Series

If \(\displaystyle \sum_{n \in \mathcal{D}} a_n\) converges to \(L\), we write

\[ \sum_{n \in \mathcal{D}} a_n = L \]

Instead of calling \(L\) the limit of \(\displaystyle \sum_{n \in \mathcal{D}} a_n\), we usually say it is its value.

Example: Geometric Series

A geometric series is a real series

\[\sum_{n = s}^{\infty} q^n,\]

where \(q\) is some real number and \(s \in \mathbb{N}_0\).

If \(|q| \lt 1\), then the geometric series is convergent with

\[\sum_{n = s}^{\infty} q^n = \frac{q^s}{1 - q}.\]

If \(|q| \ge 1\), then the geometric series is divergent.

Example: The Harmonic Series

The harmonic series is the following real series:

\[\sum_{n = 1}^{\infty} \frac{1}{n}\]

It diverges towards \(+\infty\):

\[\sum_{n = 1}^{\infty} \frac{1}{n} = +\infty\]

Definition: Absolute Convergence of Real Series

A real series \(\displaystyle \sum_{n \in \mathcal{D}} a_n\) converges absolutely or is absolutely convergent if the series \(\displaystyle \sum_{n \in \mathcal{D}} |a_n|\) converges.

Theorem: Alternative Definition

A real series \(\displaystyle \sum_{n \in \mathcal{D}} a_n\) is absolutely convergent if and only if \(\displaystyle \sum_{n \in \mathcal{D}} |a_n|\) is bounded.

Proof

TODO

Theorem: Absolute Convergence \(\implies\) Convergence

If \(\sum_{n \in \mathcal{D}} |a_n|\) converges to \(L \in \mathbb{R}\), then \(\sum_{n \in \mathcal{D}} a_n\) also converges to \(L\).

Proof

TODO

Definition: Conditional Convergence

A real series is conditionally convergent iff it is convergent but not absolutely convergent.

Example: The Alternating Harmonic Series

The alternating harmonic series is the following real series:

\[\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots\]

It converges towards the natural logarithm of \(2\):

\[\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} = \ln 2\]

However, it is not absolutely convergent because \(\sum_{n = 1}^{\infty} \left\vert \frac{(-1)^{n+1}}{n}\right\vert = \sum_{n = 1}^{\infty} \frac{1}{n}\). This is the harmonic series and we know that the harmonic series is divergent. Therefore, \(\sum_{n = 1}^{\infty} \frac{(-1)^{n+1}}{n}\) is conditionally convergent.

Theorem: Arithmetic with Real Series

If \(\sum_{n \in \mathcal{D}} a_n\) and \(\sum_{n \in \mathcal{D}} b_n\) are convergent real series, then

\[ \sum_{n \in \mathcal{D}} (\lambda a_n + \mu b_n) = \lambda \sum_{n \in \mathcal{D}} a_n + \mu \sum_{n \in \mathcal{D}} b_n \]

for all \(\lambda, \mu \in \mathbb{R}\).

Proof

TODO

The Riemann Series Theorem

Let \(\sum_{n \in \mathcal{D}} a_n\) be an infinite real series.

Definition: Rearrangement

A rearrangement of \(\sum_{n \in \mathcal{D}} a_n\) is another infinite real series \(\sum_{n \in \mathcal{D}} {}_{\sigma}a_n\) such that \({}_{\sigma} a_n\) can be expressed as

\[ {}_{\sigma}a_{n} = a_{\sigma (n)} \]

for some permutation \(\sigma\) of \(\mathcal{D}\).

If \(\sum_{n \in \mathcal{D}} a_n\) is conditionally convergent, then for each \(L \in \mathbb{R}\), there exists a rearrangement of \((a_n)_{n \in \mathcal{D}}\) which converges to \(L\). There is also a rearrangement which diverges to \(+\infty\), a rearrangement which diverges to \(-\infty\) and a rearrangement which simply diverges.

Intuition

The terms of every conditionally convergent real series can also be rearranged to produce a sum which is equal to any real number of our choice, or which diverges to infinity or negative infinite or just does not approach any limit, finite or infinite.

Proof

TODO

Theorem: Convergent Series \(\implies\) Zero Sequence

If an infinite real series \(\sum_{n \in \mathcal{D}} a_n\) is convergent, then \((a_n)_{n \in \mathcal{D}}\) is a zero sequence.

Proof

Since the real series is convergent, the Cauchy convergence theorem tells us that for each \(\varepsilon \gt 0\), there exists some \(N_{\varepsilon} \in \mathcal{D}\) such that

\[ \left\vert \sum_{k = m + 1}^n a_k \right\vert \lt \varepsilon \]

for all \(n,m \in \mathcal{D}\) with \(n,m \ge N_{\varepsilon}\) and \(n \gt m\).

If we set \(m = n - 1\), we get

\[ \left\vert \sum_{k = m + 1}^n a_k \right\vert = \left\vert \sum_{k = n}^n a_k \right \vert = |a_n| = |a_n - 0| \lt \varepsilon. \]

This holds for all \(n \ge N_{\varepsilon}\), i.e. it is the definition of the limit of a real sequence.

Theorem: Minorant Divergence Test

Let \(\displaystyle \sum_{n \in \mathcal{D}} a_n\) be a real series.

If there exists some divergent real series \(\displaystyle \sum_{n \in \mathcal{D}} b_n\) and some integer \(N\) such that \(0 \le b_n \le a_n\) for all \(n \ge N\), then \(\displaystyle \sum_{n \in \mathcal{D}} a_n\) also diverges.

Proof

TODO

Example: The Series \(\sum_{n = 1}^{\infty} \frac{1}{n^{\alpha}}\)

If \(\alpha \le 1\), then the real series \(\sum_{n = 1}^{\infty} \frac{1}{n^{\alpha}}\) diverges, since

\[\frac{1}{n} \le \frac{1}{n^{\alpha}} \qquad \forall n \in \mathbb{N}\]

and the harmonic series \(\sum_{n = 1}^{\infty} \frac{1}{n}\) diverges.

Theorem: Cauchy Convergence Theorem

A real series \(\sum_{n \in \mathcal{D}} a_n\) is convergent if and only if for each \(\varepsilon \gt 0\), there exists some \(N_{\varepsilon} \in \mathcal{D}\) such that

\[\left\vert \sum_{k = m + 1}^n a_k \right\vert \lt \varepsilon\]

for all \(m,n \in \mathcal{D}\) with \(n \gt m\) and \(m,n \ge N_{\varepsilon}\).

Proof

TODO

Theorem: Majorant Convergence Test

Let \(\displaystyle \sum_{n \in \mathcal{D}} a_n\) be a real series.

If there exists some convergent real series \(\displaystyle \sum_{n \in \mathcal{D}} b_n\) and some integer \(N\) such that \(|a_n| \le b_n\) for all \(n \ge N\), then \(\displaystyle \sum_{n \in \mathcal{D}} a_n\) is absolutely convergent.

Proof

TODO

Example: The Series \(\sum_{n = 1}^{\infty} \frac{1}{n^{\alpha}}\)

If \(\alpha \gt 1\), then the real series \(\sum_{n = 1}^{\infty} \frac{1}{n^{\alpha}}\) is convergent.

TODO

Theorem: Ratio Test

Let \(\sum_{n \in \mathcal{D}} a_n\) be an infinite real series.

(I) If there exists some \(N \in \mathcal{D}\) and some \(q \in (0;1)\) such that \(\left\vert \frac{a_{n+1}}{a_n}\right\vert \le q\) for all \(n \ge N\), then \(\sum_{n \in \mathcal{D}} a_n\) is absolutely convergent.
(II) If there exists some \(N \in \mathcal{D}\) such that \(\left\vert \frac{a_{n+1}}{a_n}\right\vert \ge 1\) for all \(n \ge N\), then \(\sum_{n \in \mathcal{D}} a_n\) is divergent.

Proof

Proof of (I):

Pick any \(q \lt 1\) such that \(\left\vert \frac{a_{n+1}}{a_n}\right\vert \le q\) for all \(n \ge N\). We notice that

\[ |a_{n+1}| \le q|a_n| \le q^2 |a_{n-1}| \le \cdots, \]

i.e. \(|a_n| \le q^{n - N} |a_N|\) for all \(n \ge N\). The series \(\sum_{n = N}^{\infty} q^{n - N}|a_N|\) is the same as \(\sum_{n = 0}^{\infty} q^n |a_N|\), which is convergent because \(|a_N|\) is just a real number and the geometric series \(\sum_{n = 0}^{\infty} q^n\) is convergent for \(q \lt 1\). Therefore, the majorant convergence test tells us that \(\sum_{n \in \mathcal{D}} |a_n|\) converges, i.e. \(\sum_{n \in \mathcal{D}} a_n\) is absolutely convergent.

Proof of (II):

We show that \((a_n)\) is not a zero sequence and so \(\sum_{n \in \mathcal{D}}\) must diverge.

Suppose \((a_n)\) is a zero sequence:

\[\lim_{n \to \infty} a_n = 0\]

Theorem: Ratio Test via Limits

If \(\left\vert\frac{a_{n+1}}{a_n}\right\vert\) is convergent, then the ratio test can be alternatively stated in the following way:

If \(\lim_{n \in \mathcal{D}} \left\vert\frac{a_{n+1}}{a_n}\right\vert \lt 1\), then \(\sum_{n \in \mathcal{D}} a_n\) is absolutely convergent.
If \(\lim_{n \in \mathcal{D}} \left\vert\frac{a_{n+1}}{a_n}\right\vert \gt 1\), then \(\sum_{n \in \mathcal{D}} a_n\) is divergent.
If \(\lim_{n \in \mathcal{D}} \left\vert\frac{a_{n+1}}{a_n}\right\vert = 1\), then we can't use this limit to make any statements about \(\sum_{n \in \mathcal{D}} a_n\).

Proof

TODO

Example: The Series \(\sum_{n = 0}^{\infty} \frac{1}{n!}\)

The real series \(\sum_{n = 0}^{\infty} \frac{1}{n!}\) is absolutely convergent, since

\[\left\vert\frac{a_{n+1}}{a_n}\right\vert = \frac{n!}{(n+1)!} = \frac{1}{n+1}\]

and

\[\lim_{n \to \infty} \frac{1}{n+1} = 0 \lt 1.\]

Example: The Series \(\sum_{n=1}^{\infty} \frac{n^2}{3^n}\)

The real series \(\sum_{n=1}^{\infty} \frac{n^2}{3^n}\) is absolutely convergent, since

\[\left\vert\frac{a_{n+1}}{a_n}\right\vert = \frac{(n+1)^2}{3^{n+1}}\frac{3^n}{n^2} = \frac{1}{3}\frac{n^2 + 2n + 1}{n^2}\]

and

\[\lim_{n \to \infty} \frac{1}{3}\frac{n^2 + 2n + 1}{n^2} = \frac{1}{3} \lt 1.\]

Theorem: \(n\)-th Root Convergence Test

Let \(\sum_{n \in \mathcal{D}} a_n\) be a real series.

If the real sequence \(\sqrt[n]{|a_n|}\) converges, then \(\sum_{n \in \mathcal{D}} a_n\) is:

absolutely convergent if

\[\displaystyle \lim_{n \to \infty} \sqrt[n]{|a_n|} \lt 1\]

divergent if

\[\displaystyle \lim_{n \to \infty} \sqrt[n]{|a_n|} \gt 1\]

Note

If \(\displaystyle \lim_{n \to \infty} \sqrt[n]{|a_n|} = 1\), then this theorem is useless.

Proof

TODO

Example: The Series \(\sum_{n=3}^{\infty} \left(1 - \frac{2}{n}\right)^{n^2}\)

The real sequence \(\sum_{n=3}^{\infty} \left(1 - \frac{2}{n}\right)^{n^2}\) is absolutely convergent, since

\[\sqrt[n]{|a_n|} = \left(1 - \frac{2}{n}\right)^n\]

and the sequence

\[\left(1 - \frac{2}{n}\right)^n\]

converges to \(\mathrm{e}^{-2}\).

Theorem: Leibniz Convergence Test for Alternating Series

Let \(\sum_{n \in \mathcal{D}} (-1)^n a_n\) be an infinite real series.

If \(a_n \ge 0\) for all \(n \in \mathcal{D}\) and \((a_n)_{n \in \mathcal{D}}\) is a decreasing zero sequence, then \(\sum_{n \in \mathcal{D}} (-1)^n a_n\) converges.

If \(a_n \le 0\) for all \(n \in \mathcal{D}\) and \((a_n)_{n \in \mathcal{D}}\) is an increasing zero sequence, then \(\sum_{n \in\mathcal{D}} (-1)^n a_n\) converges.

Proof

TODO

Example: The Series \(\sum_{n = 1}^{\infty} \frac{(-1)^n}{n^{\alpha}}\)

Consider the real series

\[\sum_{n = 1}^{\infty} \frac{(-1)^n}{n^{\alpha}},\]

where \(\alpha \in \mathbb{R}\).

If \(\alpha \gt 0\), then the series is convergent because \(\frac{1}{n^{\alpha}}\) is a decreasing sequence which converges to \(0\).

Theorem: The Integral Test

A real series

\[\sum_{n = a}^{\infty} f(n),\]

where \(f: [a, +\infty) \subset \mathbb{R} \to \mathbb{R}\) is a non-negative decreasing real function, is convergent if and only if the improper integral

\[\int_a^{\infty} f(x) \,\mathrm{d}x\]

is [convergent](./Real%20Functions/Limits%20(Real%20Functions.md).

Example: \(\sum_{n=2}^{\infty} \frac{1}{n \ln^{\alpha} n}\)

We want to see for what values of \(\alpha\) the following real series converges:

\[\sum_{n=2}^{\infty} \frac{1}{n \ln^{\alpha} n}\]

We examine the corresponding improper integral:

\[\int_2^{\infty} \frac{1}{x \ln^{\alpha}x}\,\mathrm{d}x = \lim_{\beta \to \infty} \int_2^{\beta} \frac{1}{x \ln^{\alpha}x}\,\mathrm{d}x\]

Using integration by substitution with \(u = \ln x\):

\[\int_2^{\beta} \frac{1}{x \ln^{\alpha}x}\,\mathrm{d}x = \int_{\ln 2}^{\ln \beta} \frac{1}{u^{\alpha}} \,\mathrm{d}u\]

We know that

\[\lim_{\beta \to \infty} \int_2^{\beta} \frac{1}{u^{\alpha}}\,\mathrm{d}u = \int_2^{\infty} \frac{1}{u^{\alpha}}\,\mathrm{d}u\]

[converges](./Real%20Functions/Limits%20(Real%20Functions.md) if and only if \(\alpha \gt 1\) and so the original improper integral:

\[\int_2^{\infty} \frac{1}{x \ln^{\alpha}x}\,\mathrm{d}x\]

[converges](./Real%20Functions/Limits%20(Real%20Functions.md) if and only if \(\alpha \gt 1\).

Proof

TODO