Checking Decrease and the Zero Limit

Why Check the Magnitudes?

An alternating series changes sign from term to term, but its convergence depends on more than the sign pattern. The size of each term must also behave in a particular way: the magnitudes must not increase, and they must tend to zero. For an explicit series, those conditions become concrete questions about a sequence \(b_n\) of positive numbers.

This tutorial focuses on verifying those conditions carefully. Two common approaches are to compare consecutive terms directly and to extend the sequence to a differentiable function whose derivative controls its behavior. In either approach, checking that the terms decrease and checking their limit are separate tasks.

Alternating-series conditions: For a series \(\sum_{n=1}^{\infty}(-1)^{n-1}b_n\), the alternating-series test applies if \(b_n>0\) for every \(n\), \(b_{n+1}\le b_n\) for every \(n\), and \(\lim_{n\to\infty}b_n=0\). Under these conditions, the alternating series converges.

The inequality \(b_{n+1}\le b_n\) says that the sequence is nonincreasing; equality is allowed. If \(b_{n+1}<b_n\) for every \(n\), it is strictly decreasing, which is more than the test requires. Neither form of decrease, by itself, guarantees a zero limit.

Compare Consecutive Terms Directly

When the formula is manageable, the most direct verification is to calculate \(b_n-b_{n+1}\). If \(b_n\) and \(b_{n+1}\) are positive and the difference is nonnegative for every relevant integer \(n\), then \(b_{n+1}\le b_n\). Alternatively, for positive terms, one can compare the ratio \(b_{n+1}/b_n\) with \(1\). In either calculation, denominators must be known to be positive before multiplying or dividing by them.

Worked Example: A Rational Magnitude Sequence

Consider \(b_n=\frac{n}{n^2+1}\), for \(n\in\mathbb N\). First, \(n>0\) and \(n^2+1>0\), so \(b_n>0\). To check decrease, calculate the consecutive difference:

\begin{aligned} b_n-b_{n+1} &=\frac{n}{n^2+1}-\frac{n+1}{(n+1)^2+1}\\ &=\frac{n\big((n+1)^2+1\big)-(n+1)(n^2+1)} {(n^2+1)\big((n+1)^2+1\big)}\\ &=\frac{n(n^2+2n+2)-(n+1)(n^2+1)} {(n^2+1)\big((n+1)^2+1\big)}\\ &=\frac{n^3+2n^2+2n-(n^3+n^2+n+1)} {(n^2+1)\big((n+1)^2+1\big)}\\ &=\frac{n^2+n-1}{(n^2+1)\big((n+1)^2+1\big)}. \end{aligned}

The denominator is positive. Since \(n\ge1\), the numerator satisfies \(n^2+n-1\ge1+1-1=1>0\). Thus \(b_n-b_{n+1}>0\), so the sequence is strictly decreasing. For the limit, divide the numerator and denominator of \(b_n\) by \(n^2\):

b_n=\frac{1/n}{1+1/n^2}\longrightarrow\frac{0}{1}=0.

Both magnitude conditions hold, and \(b_n\) is positive. Therefore \(\sum_{n=1}^{\infty}(-1)^{n-1}\frac{n}{n^2+1}\) satisfies the alternating-series test and converges. As a check on the first comparison, \(b_1=\frac12\) and \(b_2=\frac25\), and \(\frac25<\frac12\), as the general inequality predicts.

Use a Derivative to Check Decrease

A sequence can often be written as \(b_n=f(n)\) for a function \(f\) defined on the real interval \([1,\infty)\). If \(f\) is differentiable and \(f'(x)\le0\) throughout that interval, the mean value theorem shows that \(f\) is nonincreasing there. Consequently, its values at successive integers satisfy \(f(n+1)\le f(n)\), which is exactly \(b_{n+1}\le b_n\).

Derivative criterion: Suppose \(f\) is continuous on every interval \([n,n+1]\), differentiable on \((n,n+1)\), and \(f'(x)\le0\) for \(x\in(n,n+1)\). The mean value theorem gives \(f(n+1)-f(n)=f'(c)\) for some \(c\in(n,n+1)\). Hence \(f(n+1)\le f(n)\). If this holds for each \(n\in\mathbb N\), then \(b_n=f(n)\) is nonincreasing.

The criterion concerns the derivative over the entire interval between consecutive integers, not just at the integer values. Once decrease is established, the limit still needs its own calculation. If \(f(x)\to0\) as \(x\to\infty\), then \(b_n=f(n)\to0\), because the sequence takes values of the function along inputs that grow without bound.

Worked Example: A Logarithmic Magnitude Sequence

Let \(b_n=\frac{\ln(n+1)}{n}\), for \(n\in\mathbb N\). Define \(f(x)=\frac{\ln(x+1)}{x}\) for \(x\ge1\), so \(b_n=f(n)\). The logarithm is positive when \(x\ge1\), and \(x>0\), so \(f(x)>0\). Differentiating gives

f'(x)=\frac{\frac{x}{x+1}-\ln(x+1)}{x^2}.

To establish the sign of the numerator, define \(h(x)=\ln(x+1)-\frac{x}{x+1}\) for \(x\ge0\). Then \(h(0)=0\), and

h'(x)=\frac{1}{x+1}-\frac{1}{(x+1)^2} =\frac{x}{(x+1)^2}>0\qquad(x>0).

By the mean value theorem, \(h\) is strictly increasing on every interval in \([0,\infty)\). Thus \(h(x)>h(0)=0\) for \(x>0\). It follows that \(\frac{x}{x+1}-\ln(x+1)=-h(x)<0\), and therefore \(f'(x)<0\) for \(x\ge1\). The sequence \(b_n\) is positive and strictly decreasing.

For the limit, apply l’Hôpital’s rule to the continuous-variable quotient as \(x\to\infty\). Its numerator and denominator both tend to infinity, and their derivatives are \(\frac1{x+1}\) and \(1\), respectively:

\lim_{x\to\infty}\frac{\ln(x+1)}{x} =\lim_{x\to\infty}\frac{1/(x+1)}{1}=0.

Hence \(b_n\to0\). The positive magnitudes satisfy both required conditions, so \(\sum_{n=1}^{\infty}(-1)^{n-1}\frac{\ln(n+1)}{n}\) converges by the alternating-series test. For example, its first magnitudes are \(b_1=\ln 2\) and \(b_2=\frac{\ln3}{2}\); the derivative argument establishes the comparison for every consecutive pair, not just these first terms.

A Direct Comparison and Limit

Some sequences are easiest to check by their elementary structure. If a positive denominator increases, its reciprocal decreases; if a positive quantity increases, its reciprocal square root also decreases. It is still useful to write the comparison explicitly, so the direction of the inequality is unmistakable.

Worked Example: Reciprocal Square-Root Magnitudes

Take \(b_n=\frac1{\sqrt n}\), for \(n\in\mathbb N\). Each term is positive. Since \(n+1>n>0\), taking square roots preserves the inequality: \(\sqrt{n+1}>\sqrt n>0\). Taking reciprocals of positive quantities reverses it, so

b_{n+1}=\frac1{\sqrt{n+1}}<\frac1{\sqrt n}=b_n.

Thus \(b_n\) is strictly decreasing. Also, \(n\to\infty\) implies \(\sqrt n\to\infty\), so its reciprocal tends to zero:

\lim_{n\to\infty}b_n=\lim_{n\to\infty}\frac1{\sqrt n}=0.

Therefore, the alternating series \(\sum_{n=1}^{\infty}(-1)^{n-1}\frac1{\sqrt n}\) meets the alternating-series test. The initial comparison checks numerically as well: \(b_1=1\) and \(b_2=\frac1{\sqrt2}<1\).

Why Decrease Alone Is Not Enough

A strictly decreasing positive sequence can settle toward a positive number rather than zero. The zero-limit condition is therefore not a consequence of the decrease condition; it must be checked independently. This distinction matters because a necessary first check for any convergent series is that its terms tend to zero.

Worked Example: Decrease Without a Zero Limit

Consider \(b_n=1+\frac1n\). Every term is positive, and

b_n-b_{n+1} =\left(1+\frac1n\right)-\left(1+\frac1{n+1}\right) =\frac1n-\frac1{n+1} =\frac{1}{n(n+1)}>0.

So \(b_n\) is strictly decreasing. However,

\lim_{n\to\infty}b_n =1+\lim_{n\to\infty}\frac1n =1\ne0.

The alternating-series test does not apply. In fact, the terms \((-1)^{n-1}b_n\) do not tend to zero: their magnitudes tend to \(1\). Thus the series cannot converge.

A Reliable Verification Procedure

For a proposed alternating series \(\sum_{n=1}^{\infty}(-1)^{n-1}b_n\), organize the check so each condition receives its own justification. A derivative calculation is a tool for the inequality, not a substitute for the limit calculation.

Check positivity.
Verify from the formula that \(b_n>0\) for every positive integer \(n\).

Check consecutive magnitudes.
Prove \(b_{n+1}\le b_n\) directly, or write \(b_n=f(n)\) and establish \(f'(x)\le0\) on the intervals between successive integers.

Check the limit separately.
Evaluate \(\lim_{n\to\infty}b_n\) using algebra, a comparison, or an applicable limit rule, and verify that it equals zero.

State the conclusion precisely.
Only after all the conditions hold should you invoke the alternating-series test to conclude convergence.

Common Pitfall: Confusing Decrease with Convergence to Zero

A common incorrect argument is: “The magnitudes are positive and decreasing, so they approach zero.” The example \(b_n=1+\frac1n\) disproves that claim: its terms decrease, but their limit is \(1\). The correction is to prove the consecutive-term inequality and compute the limit as two distinct steps. Conversely, finding \(b_n\to0\) does not prove the terms decrease; that inequality also needs evidence.

When using derivatives, another avoidable error is to calculate \(f'(n)\) for integer inputs and treat its sign there as a complete proof of the consecutive inequalities. The derivative criterion requires a sign throughout each interval \((n,n+1)\). Establishing \(f'(x)\le0\) for every \(x\ge1\) supplies that interval-wide evidence; the mean value theorem then connects it to the integer sequence.

Key takeaway: For alternating-series magnitudes, verify positivity, nonincrease, and a zero limit separately. A direct consecutive comparison proves the discrete inequality; a derivative with the right sign proves it through the mean value theorem. Neither method removes the need to calculate the limit.

Check Your Understanding

For each question, identify which magnitude condition is being checked and give a justification.

For \(b_n=\frac{1}{n+2}\), show directly that \(b_{n+1}<b_n\) for every \(n\in\mathbb N\), and evaluate its limit.
Suppose \(b_n=f(n)>0\), where \(f\) is differentiable on \([1,\infty)\) and \(f'(x)\le0\) for all \(x>1\). Explain why \(b_{n+1}\le b_n\).
For \(b_n=\frac{n+1}{n}\), determine whether the sequence decreases and whether it tends to zero. Does it meet the alternating-series test’s magnitude conditions?
Why does verifying \(b_{n+1}\le b_n\) for several initial values not establish that inequality for every positive integer \(n\)?
For \(b_n=\frac{1}{\sqrt{n+3}}\), verify positivity, decrease, and the zero limit, then state what the alternating-series test concludes for \(\sum_{n=1}^{\infty}(-1)^{n-1}b_n\).