Divergence-Comparison Workshop: Harmonic Lower Bounds

Why a Lower Bound Can Prove Divergence

A positive series can diverge because its terms do not become small quickly enough. To establish divergence, it is often useful to compare each term with a familiar divergent series. The key direction is a lower bound: if the terms of the series under consideration are at least as large as terms whose sum is already known to diverge, then their partial sums cannot remain bounded.

In this workshop, the comparison series is the harmonic series, whose terms are \(1/n\). The goal is to find a constant \(c>0\) and an index \(N\) such that the terms \(a_n\) satisfy \(a_n\ge c/n\) for every \(n\ge N\). The constant need not be large. Any fixed positive multiple of the harmonic terms still has divergent partial sums.

Definition: A positive series is a series \(\sum_{n=1}^{\infty} a_n\) whose terms satisfy \(a_n>0\) for every index under consideration. Its \(k\)-th partial sum is \(S_k=\sum_{n=1}^{k}a_n\). The series diverges to infinity if its partial sums grow without bound.

The Harmonic Series as a Comparison Baseline

The harmonic series is \(\sum_{n=1}^{\infty}1/n\). We will use its divergence as the baseline for our comparisons. A direct proof comes from grouping terms in blocks whose endpoints are powers of two.

For each integer \(k\ge0\), consider the block of indices from \(2^k\) through \(2^{k+1}-1\). This block has \(2^k\) terms. Every index in it is less than \(2^{k+1}\), so every reciprocal in the block is greater than \(1/2^{k+1}\). Consequently, the sum in each block is greater than \(1/2\).

\sum_{n=2^k}^{2^{k+1}-1}\frac{1}{n} > 2^k\cdot\frac{1}{2^{k+1}} = \frac{1}{2}.

The blocks with \(k=0,1,\ldots,m-1\) cover exactly the indices from \(1\) through \(2^m-1\). Adding their lower bounds gives:

\sum_{n=1}^{2^m-1}\frac{1}{n} = \sum_{k=0}^{m-1}\sum_{n=2^k}^{2^{k+1}-1}\frac{1}{n} > \frac{m}{2}.

As \(m\) increases, \(m/2\) grows without bound. The harmonic partial sums therefore grow without bound, so the harmonic series diverges.

The Divergence Form of Direct Comparison

Theorem: Suppose \(a_n\ge0\) and \(b_n\ge0\) for all \(n\ge N\), and \(a_n\ge b_n\) for every \(n\ge N\). If \(\sum_{n=N}^{\infty}b_n\) diverges to infinity, then \(\sum_{n=1}^{\infty}a_n\) diverges to infinity. In particular, if \(a_n\ge c/n\) for some constant \(c>0\) and every \(n\ge N\), then \(\sum a_n\) diverges.

The finite number of terms before \(N\) does not affect whether the partial sums grow without bound. The essential comparison is between the tails.

Why the Comparison Rule Works

Let \(T_k=\sum_{n=N}^{k}b_n\) and \(U_k=\sum_{n=N}^{k}a_n\), where \(k\ge N\). Since each term in the second sum is at least the corresponding term in the first, adding the inequalities gives \(U_k\ge T_k\). If the partial sums \(T_k\) grow without bound, then \(U_k\) must also grow without bound. Adding the fixed initial sum \(\sum_{n=1}^{N-1}a_n\) cannot change that conclusion.

\sum_{n=1}^{k}a_n = \sum_{n=1}^{N-1}a_n+\sum_{n=N}^{k}a_n \ge \sum_{n=1}^{N-1}a_n+\sum_{n=N}^{k}b_n.

For the particular choice \(b_n=c/n\), the tail partial sums of \(b_n\) grow without bound because \(c>0\) and multiplying the harmonic partial sums by \(c\) preserves their unbounded growth.

Worked Example: A Square-Root Denominator

Worked Example: Prove that \(\sum_{n=1}^{\infty}1/(n+\sqrt n)\) diverges

For every integer \(n\ge1\), we have \(\sqrt n\le n\). Indeed, both sides are nonnegative, and squaring gives the equivalent inequality \(n\le n^2\), which holds for \(n\ge1\). Therefore \(n+\sqrt n\le2n\). Both denominators are positive, so taking reciprocals reverses the inequality:

\frac{1}{n+\sqrt n}\ge\frac{1}{2n} =\frac12\cdot\frac{1}{n}.

The right-hand series is one-half the harmonic series, so it diverges. The terms on the left are at least as large for every \(n\ge1\); direct comparison proves that \(\sum_{n=1}^{\infty}1/(n+\sqrt n)\) diverges.

Check: At \(n=1\), the original term is \(1/(1+1)=1/2\), and the lower bound is \(1/(2\cdot1)=1/2\). The inequality is attained there. For every \(n\ge1\), the denominator inequality above verifies the comparison used in the proof.

Worked Example: A Logarithmic Denominator

Worked Example: Prove that \(\sum_{n=1}^{\infty}1/(n+\ln n)\) diverges

For \(n\ge1\), the logarithm is nonnegative. Also, using \(\ln n=\int_1^n 1/x\,dx\) and \(1/x\le1\) on \([1,n]\), we obtain \(\ln n\le n-1\). Thus \(n+\ln n\le2n-1\le2n\). The denominator is positive, so:

\frac{1}{n+\ln n} \ge \frac{1}{2n} = \frac12\cdot\frac{1}{n}.

The harmonic comparison diverges, and the original terms are positive and no smaller than half the corresponding harmonic terms. Therefore \(\sum_{n=1}^{\infty}1/(n+\ln n)\) diverges.

Check: At \(n=1\), \(\ln 1=0\), so the original term is \(1\), while the stated lower bound is \(1/2\). For general \(n\ge1\), the bound \(\ln n\le n-1\) gives \(n+\ln n\le2n-1\), which confirms that the denominator is no larger than \(2n\) and hence the reciprocal is no smaller than \(1/(2n)\).

Worked Example: A Bound That Starts Later

Worked Example: Prove that \(\sum_{n=1}^{\infty}1/(n+\sqrt n+5)\) diverges

Here the constant \(5\) means the simple bound by \(3n\) is not useful at every index. For \(n\ge5\), we have \(\sqrt n\le n\) and \(5\le n\). Adding these inequalities to the \(n\) already in the denominator yields:

n+\sqrt n+5\le n+n+n=3n \qquad(n\ge5).

All terms in the denominator are positive, so taking reciprocals gives the eventual lower bound:

\frac{1}{n+\sqrt n+5}\ge\frac{1}{3n} =\frac13\cdot\frac{1}{n} \qquad(n\ge5).

The tail \(\sum_{n=5}^{\infty}1/(3n)\) diverges, since deleting the first four terms from a divergent harmonic series does not make its partial sums bounded, and multiplying the terms by \(1/3\) preserves unboundedness. Direct comparison proves divergence of the original series. Its first four terms are finite, so their inclusion does not alter the conclusion.

Check: At the starting index \(n=5\), the denominator is \(5+\sqrt5+5=10+\sqrt5\), which is less than \(15=3\cdot5\) because \(\sqrt5<5\). Thus the term exceeds \(1/15\), as required. For all \(n\ge5\), the two inequalities \(\sqrt n\le n\) and \(5\le n\) establish the claimed comparison.

A Reusable Proof Plan

For a denominator that includes a positive main term and smaller additions, the useful task is often to show that the whole denominator is at most a constant times \(n\). The reciprocal then provides the required harmonic lower bound. Keep the direction of the inequalities explicit at each step.

Check positivity.
Verify that the series terms and denominators are positive on the indices where the comparison will be made.

Bound the denominator above.
Find a constant \(C>0\) and an index \(N\) such that the denominator is at most \(Cn\) for all \(n\ge N\).

Take reciprocals carefully.
For positive quantities, a smaller denominator gives a larger reciprocal, so the term is at least \(1/(Cn)\).

Apply divergence comparison.
Since \(1/(Cn)\) is a positive constant multiple of the harmonic term, the series diverges.

Common Pitfall: Reversing the Comparison Direction

A lower bound on a series term is what proves divergence by comparison with a divergent series. For example, the following calculation is wrong:

n+\sqrt n\le2n \quad\Longrightarrow\quad \frac{1}{n+\sqrt n}\le\frac{1}{2n}.

The implication reverses incorrectly. Because both denominators are positive, taking reciprocals reverses the inequality. The correct calculation is:

n+\sqrt n\le2n \quad\Longrightarrow\quad \frac{1}{n+\sqrt n}\ge\frac{1}{2n}.

Another important check is whether the bound holds for the stated indices. If it only holds for \(n\ge N\), use the divergence comparison on that tail and note that the omitted initial terms form a finite sum. Do not silently claim an inequality for indices where it has not been verified.

Why These Bounds Are Useful

The comparisons here do not require a precise estimate of every term. They show that the terms remain at least as large as a fixed fraction of \(1/n\), which is enough to force the partial sums to grow without bound. In particular, a square-root or logarithmic addition to \(n\) does not prevent the denominator from being bounded above by a constant multiple of \(n\) in these examples.

The practical discipline is to justify the denominator bound first, check positivity, and only then take reciprocals. This keeps the inequality direction clear and makes the role of the starting index explicit. Once the lower bound \(a_n\ge c/n\) has been established for all sufficiently large \(n\), the divergence conclusion follows directly from the harmonic comparison.

Check Your Understanding

For each question, identify the index range and inequality needed for the comparison.

For \(n\ge1\), show that \(n+\sqrt n\le2n\), and state the resulting lower bound for \(1/(n+\sqrt n)\).
Use \(\ln n\le n-1\) for \(n\ge1\) to find a constant-multiple harmonic lower bound for \(1/(n+\ln n)\).
For \(n\ge7\), find a constant \(C\) such that \(n+\sqrt n+7\le Cn\), and state the resulting comparison.
Explain why \(a_n\ge c/n\) with \(c>0\) proves divergence when it holds only for \(n\ge N\).
Correct the inequality direction: if \(0<u\le v\), how do \(1/u\) and \(1/v\) compare?