Bounded series and convergence – "Math for Non-Geeks"

Bounded series with positive summands converge

As we learned in the chapter "monotonic functions", monotonicity and boundedness combined are a sufficient criterion to show convergence of a sequence. This criterion is also applicable to series, since the meaning of a series is defined by the sequence of its partial sums. In the following, we will derive a convergence theorem for series that is based on the monotonicity and boundedness criterion for sequences.

We proceed in two steps by separately examining the monotonicity and boundedness of partial sums $s_n\equiv \sum _{k=1}^n{a}_k$ of some series $\sum _{k=1}^{\mathrm{\infty }}{a}_k$.

First, we ask: Under what circumstances is the sequence $(s_n{)}_{n\in \mathbb{N}}$ of partial sums monotonically increasing? Starting with an arbitrary partial sum, we obtain the next one by adding the corresponding term of the series. Therefore, the next partial sum is greater (or equal), whenever this summand is positive (or zero). If we require monotonicity for the partial sums, this means all terms of the series must be non-negative (except for the first one, which does not play the role of a difference between partial sums).

We now want to show this statement formally. The relation between two subsequent partial sums is

$s_{n+1}=\sum _{k=1}^{n+1}{a}_k=a_{n+1}+\sum _{k=1}^n{a}_k=a_{n+1}+s_n.$

The sequence $(s_n{)}_{n\in \mathbb{N}}$ is monotonically increasing, if $s_{n+1}\ge s_n$ for all $n\in \mathbb{N}$. Using the relation we just derived, we find:

$\begin{array}{cccc}& & s_{n+1}& \ge s_n\\ & ⟺& a_{n+1}+s_n& \ge s_n\\ & ⟺& a_{n+1}& \ge 0\end{array}$

Hence, the sequence of partial sums is monotonically increasing, if and only if $a_{n+1}\ge 0$ for all $n\in \mathbb{N}$. This simply means $a_n\ge 0$ for all $n\ge 2$.

Next, we address boundedness of the sequence $(s_n{)}_{n\in \mathbb{N}}$. A real sequence is bounded, if there exists some constant $b\in ℝ$, such that $s_n\le b$ for all $n\in \mathbb{N}$. With the $s_n$ being partial sums, we can picture this in the following way: Each partial sum $s_n$ is a truncation of the infinite sum $\sum _{k=1}^{\mathrm{\infty }}{a}_k$, where we drop all terms after $a_n$. Each partial sum is just a finite sum of finite terms, so it is clearly finite itself. Therefore, it is not hard to find a common upper bound for some of the partial sums. However, to show boundedness of the sequence, we need to find an upper bound for all partial sums at once. So we are looking for a number that is greater than any of the truncated sums, no matter how late we do the truncation.

We summarize these observations in the following theorem:

Theorem (Bounded series with non-negative summands converge)

Let $\sum _{k=1}^{\mathrm{\infty }}{a}_k$ be a series with $a_k\ge 0$ for all $k\ge 2$. This series converges, if and only if there exists some bound $b\in ℝ$, such that $b\ge \sum _{k=1}^n{a}_k$ for all $n\in \mathbb{N}$.

Proof (Bounded series with non-negative summands converge)

Let $\sum _{k=1}^{\mathrm{\infty }}{a}_k$ be a series with $a_k\ge 0$ for all $k\ge 2$.

Proof step: If the sequence ${\left(\sum _{k=1}^{\mathrm{\infty }}{a}_k\right)}_{n\in \mathbb{N}}$ is bounded, the series converges.

We have shown above that for a series with non-negative summands, the sequence of partial sums is monotonically increasing. According to the monotonicity criterion for sequences, a monotonically increasing sequence with an upper bound converges. By definition, if its sequence of partial sums converges, a series is said to converge.

Proof step: If $\sum _{k=1}^{\mathrm{\infty }}{a}_k$ converges, its partial sums are bounded.

By definition, if the series $\sum _{k=1}^{\mathrm{\infty }}{a}_k$ converges, its sequence of partial sums converges. In chapter "Unbounded sequences diverge" we learned that convergent sequences are bounded.

Analogously, we can also prove a theorem for monotonically decreasing partial sums:

Theorem (Bounded series with non-positive summands converge)

Let $\sum _{k=1}^{\mathrm{\infty }}{a}_k$ be a series with $a_k\le 0$ for all $k\ge 2$. This series converges, if and only if there exists some bound $b\in ℝ$, such that $b\le \sum _{k=1}^n{a}_k$ for all $n\in \mathbb{N}$.

Hint

As we have seen, both theorems are applications of the monotonicity criterion for sequences. Therefore, the literature often refers to them as the monotonicity criterion for series.

Application: Convergence of the hyperharmonic series

Theorem (Convergence of the hyperharmonic series)

The hyperharmonic series $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^{\alpha }}$ is convergent for $\alpha >1$ (for $\alpha =1$ we obtain the harmonic series, whose divergence has been shown in the corresponding article).

How to get to the proof? (Convergence of the hyperharmonic series)

We want to prove convergence by applying the theorem shown above. All summands of the series are clearly positive, so it remains to show that there is an upper bound for the partial sums. We employ a similar trick as in showing the divergence of the harmonic series. We consider the first $2^{n+1}-1$ summands, which we split into suitable groups, such that each group can be assigned a bound. However, this time we choose upper bounds for the groups, as we eventually want to prove convergence. We use $2^{n+1}-1$ summands instead of $2^n$ because this eases finding a bound. At the end of a chain of inequalities, we arrive at the geometric series, whose convergence criterion has been shown in this article.

$\begin{array}{cc}\sum _{k=1}^{2^{n+1}-1}\frac{1}{k^{\alpha }}& =1+\frac{1}{2^{\alpha }}+\frac{1}{3^{\alpha }}+(\frac{1}{4^{\alpha }}+\frac{1}{5^{\alpha }}+\frac{1}{6^{\alpha }}+\frac{1}{7^{\alpha }})+\dots +\underset{2^n\text{ summands}}{\underbrace{(\frac{1}{(2^n{)}^{\alpha }}+\dots +\frac{1}{(2^{n+1}-1{)}^{\alpha }})}}\\ & \downarrow \frac{1}{3^{\alpha }}\le \frac{1}{2^{\alpha }}\wedge \frac{1}{7^{\alpha }}\le \frac{1}{6^{\alpha }}\le \frac{1}{5^{\alpha }}\le \frac{1}{4^{\alpha }}\wedge \dots \wedge \frac{1}{(2^{n+1}-1{)}^{\alpha }}\le \cdots \le \frac{1}{(2^n{)}^{\alpha }}(\text{as }\alpha >1)\\ & \le 1+\frac{1}{2^{\alpha }}+\frac{1}{2^{\alpha }}+(\frac{1}{4^{\alpha }}+\frac{1}{4^{\alpha }}+\frac{1}{4^{\alpha }}+\frac{1}{4^{\alpha }})+\dots +\underset{2^n\text{ summands}}{\underbrace{(\frac{1}{(2^n{)}^{\alpha }}+\dots +\frac{1}{(2^n{)}^{\alpha }})}}\\ & =1+(2\cdot \frac{1}{2^{\alpha }})+4\cdot \frac{1}{4^{\alpha }}+\cdots +2^n\cdot \frac{1}{(2^n{)}^{\alpha }}\\ & =1+\frac{1}{2^{\alpha -1}}+\frac{1}{4^{\alpha -1}}+\cdots +\frac{1}{(2^n{)}^{\alpha -1}}\\ & =\frac{1}{(2^{\alpha -1}{)}^0}+\frac{1}{(2^{\alpha -1}{)}^1}+\frac{1}{(2^{\alpha -1}{)}^2}+\cdots +\frac{1}{(2^{\alpha -1}{)}^n}\\ & =\sum _{l=0}^n\frac{1}{(2^{\alpha -1}{)}^l}\\ & =\sum _{l=0}^n{\left(\frac{1}{2^{\alpha -1}}\right)}^l\\ & \downarrow \text{supplement with infinitely many positive summands}\\ & \le \sum _{l=0}^{\mathrm{\infty }}{\left(\frac{1}{2^{\alpha -1}}\right)}^l\\ & \downarrow \begin{array}{cc}\alpha >1& \Rightarrow \alpha -1>0\\ & \Rightarrow 2^{\alpha -1}>1\\ & \Rightarrow \frac{1}{2^{\alpha -1}}<1\\ & \Rightarrow \text{geometric series converges}\end{array}\\ & =\frac{1}{1-\frac{1}{2^{\alpha -1}}}\\ & =\frac{1}{1-2^{1-\alpha }}<\mathrm{\infty }\end{array}$

Hence we have found an upper bound for the partial sum $s_{2^{n-1}+1}$. Moreover, this upper bound is independent of $n$, so it is not only an upper bound for a particular partial sum, but for all $s_{2^{n-1}+1}$ at the same time. We have thus shown that the subsequence $(s_{2^{n-1}+1}{)}_{n\in \mathbb{N}}$ is bounded. However, this is not sufficient to show boundedness of the full sequence of partial sums.

To address this problem, we need an auxiliary argument. Using the Archimedean property of $\mathbb{N}$ and Bernoulli's inequality, we were able to show that for any positive integer $n>0$ there is another integer $m>0$ such that $n<2^m$. In other words, any natural number is smaller than some power of two. Similarly, we can show that for any given $n>0$ we can also find some $m>0$ such that $n\le 2^{m+1}-1$. As the sequence of partial sums is monotonically increasing, this implies $s_n\le s_{2^{m+1}-1}$. So any partial sum is bounded by one of the partial sums for which we have derived a universal bound above. We can conclude that this bound is a bound for all partial sums, such that the criterion is applicable to the series.

Proof (Convergence of the hyperharmonic series)

For any $k\ge 1$ and $\alpha >1$, we have $\frac{1}{k^{\alpha }}>0.$ This proves monotonicity of the sequence of partial sums.

Furthermore, for any $n\in \mathbb{N}$ there is some $m\in \mathbb{N}$ such that $n\le 2^{m+1}-1$. This enables the chain of inequalities $\begin{array}{cc}{s}_n=\sum _{k=1}^n\frac{1}{k^{\alpha }}\le \sum _{k=1}^{2^{m+1}-1}\frac{1}{k^{\alpha }}& =1+\frac{1}{2^{\alpha }}+\frac{1}{3^{\alpha }}+(\frac{1}{4^{\alpha }}+\frac{1}{5^{\alpha }}+\frac{1}{6^{\alpha }}+\frac{1}{7^{\alpha }})+\dots +(\frac{1}{(2^m{)}^{\alpha }}+\dots +\frac{1}{(2^{m+1}-1{)}^{\alpha }})\\ & \le 1+\frac{1}{2^{\alpha }}+\frac{1}{2^{\alpha }}+(\frac{1}{4^{\alpha }}+\frac{1}{4^{\alpha }}+\frac{1}{4^{\alpha }}+\frac{1}{4^{\alpha }})+\dots +\underset{2^n\text{ summands}}{\underbrace{(\frac{1}{(2^n{)}^{\alpha }}+\dots +\frac{1}{(2^n{)}^{\alpha }})}}\\ & \le \sum _{l=0}^m{\left(\frac{1}{2^{\alpha -1}}\right)}^l\\ & \le \sum _{l=0}^{\mathrm{\infty }}{\left(\frac{1}{2^{\alpha -1}}\right)}^l=\frac{1}{1-2^{1-\alpha }}<\mathrm{\infty }.\end{array}$

Thus, the sequence of partial sums is also bounded. The monotonicity criterion then shows convergence of the hyperharmonic series $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^{\alpha }}$ for any $\alpha >1$.