Why Verify the Magnitudes?
An alternating series changes sign from term to term, but its signs alone do not establish convergence. The size of each term matters too. To use the Alternating Series Test, you must verify that the positive magnitudes eventually decrease and tend to zero.
In many problems, the difficult part is showing the decrease correctly. A formula may not make the comparison between consecutive terms obvious, and a derivative argument must apply on the relevant interval. This tutorial develops three reliable approaches: derivatives, successive-term ratios, and direct algebra.
The Alternating Series Test states that this series converges if \(b_n\to 0\) and the magnitudes are nonincreasing from some index onward. Strict decrease is not required. Nor must the decrease begin with the first term: a finite number of initial terms do not affect whether an infinite series converges.
Method 1: Use a Derivative
Suppose the magnitudes can be written as \(b_n=f(n)\) for a function \(f\). If \(f\) is differentiable and \(f'(x)<0\) throughout an interval containing every \(x\ge N\), then \(f\) is decreasing there. Consequently, \(b_{n+1}<b_n\) for every integer \(n\ge N\). If \(f'(x)\le 0\), the same reasoning gives nonincrease.
The derivative condition must hold across the intervals between consecutive integers, not merely at the integer inputs. The Mean Value Theorem explains why that matters: for each \(n\ge N\), there is a point \(c\in(n,n+1)\) such that \(f(n+1)-f(n)=f'(c)\). The sign of \(f'(c)\) therefore determines the sign of the difference between the consecutive terms.
Worked Example: Use a Derivative When Decrease Begins Later
Consider the alternating series with magnitudes \(b_n=\frac{n}{n^2+4}\). Find where the magnitudes decrease and check the condition needed for the Alternating Series Test.
Define \(f(x)=\frac{x}{x^2+4}\), so \(b_n=f(n)\). The denominator is positive for every real \(x\), and the quotient rule gives
For \(x>2\), the numerator \(4-x^2\) is negative and the denominator is positive. Thus \(f'(x)<0\) for \(x>2\), so \(b_{n+1}<b_n\) whenever \(n\ge 3\). The decrease is eventual, even though it does not start at the first term: \(b_1=\frac15\) and \(b_2=\frac{2}{8}=\frac14\), so \(b_2>b_1\).
Also, for \(n\ge1\), \(0<b_n=\frac{n}{n^2+4}\le\frac1n\), and \(\frac1n\to0\). Therefore \(b_n\to0\). As a numerical check of the claimed decrease at \(n=3\), \(b_3=\frac3{13}\) and \(b_4=\frac4{20}=\frac15\); indeed \(\frac3{13}>\frac15\), since \(15>13\). Both required conditions hold from \(n=3\) onward, so the Alternating Series Test applies.
Method 2: Compare Successive-Term Ratios
When \(b_n>0\), comparing \(b_{n+1}\) with \(b_n\) is equivalent to comparing their ratio with \(1\): \(b_{n+1}<b_n\) exactly when \(\frac{b_{n+1}}{b_n}<1\). This method is especially convenient when factorials, powers, or products cancel in the ratio.
If the ratios have a limit \(L<1\), then they are eventually less than \(1\). To see why, choose a number \(q\) with \(L<q<1\). By the definition of limit, for all sufficiently large \(n\), \(\frac{b_{n+1}}{b_n}<q<1\). Thus the terms eventually decrease. A ratio limit of \(1\), however, does not settle the question: the ratios might be below \(1\), above \(1\), or equal to \(1\).
Worked Example: Verify Eventual Decrease with Factorials
Let \(b_n=\frac{3^n}{n!}\), where \(n\ge1\). Determine when these positive magnitudes decrease, and verify that they tend to zero.
For each \(n\ge1\),
This ratio is less than \(1\) exactly when \(n+1>3\), or \(n>2\). Therefore \(b_{n+1}<b_n\) for every \(n\ge3\). The initial terms need not decrease: \(b_1=3\), \(b_2=\frac92\), and \(b_3=\frac{27}{6}=\frac92\).
To check the limit, for \(n\ge5\) we have \(\frac{b_{n+1}}{b_n}=\frac3{n+1}\le\frac12\). Repeatedly applying this inequality gives \(0<b_n\le b_5(\frac12)^{n-5}\) for \(n\ge5\). Since \(b_5=\frac{3^5}{5!}=\frac{243}{120}=\frac{81}{40}\) and the geometric factor tends to zero, \(b_n\to0\). As a final check of the decrease at \(n=3\), \(b_3=\frac92=\frac{36}{8}\) and \(b_4=\frac{81}{24}=\frac{27}{8}\), so \(b_3>b_4\). The magnitudes decrease from \(n=3\) onward and tend to zero; the Alternating Series Test applies.
Method 3: Subtract and Simplify
For rational expressions, it can be simpler to calculate \(b_n-b_{n+1}\) directly. If the denominators are positive, put the terms over a common denominator and check whether the resulting numerator is positive. This is an exact algebraic verification, not an estimate based on a few sample terms.
Worked Example: Prove Decrease by an Algebraic Difference
Consider \(b_n=\frac{n+1}{n^2+3n+3}\) for \(n\ge1\). Show that the magnitudes decrease, and check that they tend to zero.
Write the next term with its denominator expanded: \(b_{n+1}=\frac{n+2}{(n+1)^2+3(n+1)+3}=\frac{n+2}{n^2+5n+7}\). Both denominators are positive for \(n\ge1\), so the sign of \(b_n-b_{n+1}\) is the sign of its numerator after combining fractions:
The numerator and denominator in the final expression are positive for \(n\ge1\). Hence \(b_n-b_{n+1}>0\), proving strict decrease for every \(n\ge1\).
For the limit, divide the original numerator and denominator by \(n^2\): \(b_n=\frac{\frac1n+\frac1{n^2}}{1+\frac3n+\frac3{n^2}}\to0\). Check the first consecutive terms directly: \(b_1=\frac27\) and \(b_2=\frac3{13}\); indeed, \(\frac27>\frac3{13}\), since \(26>21\). The exact difference calculation proves the comparison for every \(n\), and the limit supplies the other condition for the Alternating Series Test.
Why Eventual Decrease Is Enough
Suppose the magnitudes decrease only after an index \(N\). The terms before \(N\) form a finite sum. Adding or removing finitely many terms changes a series sum by a finite amount, but does not change whether its sequence of partial sums converges. Therefore the Alternating Series Test needs the decrease condition only from some point onward, along with \(b_n\to0\).
This does not mean the decrease condition can be skipped. It means you must identify an index after which it is valid. A proof such as “the first several terms decrease, so they keep decreasing” is not sufficient; a general argument using a derivative, ratio, or algebraic difference is needed.
A Common Ratio Mistake
A frequent error is to divide consecutive terms incorrectly, then decide decrease from the mistaken ratio. Another is to conclude that decrease follows merely because a ratio approaches \(1\). The limit \(1\) alone gives no direction of comparison.
For a concrete example, take \(b_n=\frac{n}{n+1}\). A tempting but incorrect calculation is \(\frac{b_{n+1}}{b_n}=\frac{n+1}{n+2}<1\), as though dividing the terms only required comparing their numerators or denominators separately. But division by \(b_n\) requires multiplying by its reciprocal. The correct calculation is
Thus these terms increase, not decrease, even though their successive-term ratio tends to \(1\). The correction is to form the ratio carefully and compare its exact value with \(1\); a limit equal to \(1\) is inconclusive about monotonicity.
A Reliable Verification Routine
When checking magnitudes in an alternating-series problem, first name the positive term \(b_n\) and confirm that it represents the magnitudes. Then choose a method suited to its form. In each case, state the range of indices for which the argument works.
Confirm that the quantities being compared are nonnegative; ratio comparisons require \(b_n>0\).
Use a derivative on a full interval, a correctly computed ratio, or the sign of \(b_n-b_{n+1}\).
If the argument applies only for \(n\ge N\), say so. Eventual decrease is sufficient.
Decrease alone does not establish that \(b_n\to0\); verify the limit as its own condition.
These methods turn a visual impression of “getting smaller” into a precise argument. The derivative method uses the behavior of a function between integer inputs; the ratio method turns comparison into a check against \(1\); and direct subtraction reduces the question to the sign of an expression. Each is useful when its assumptions are satisfied.
Check Your Understanding
For each question, identify the condition that must be checked and explain what conclusion the calculation supports.
- If \(b_n=f(n)\) and \(f'(x)<0\) for every \(x\ge4\), for which integer indices does the derivative argument establish decrease?
- If \(b_n>0\) and \(\frac{b_{n+1}}{b_n}=\frac{2}{n+1}\), from what index onward are the magnitudes strictly decreasing?
- Why can a finite number of increases at the start of an alternating series be consistent with applying the Alternating Series Test?
- For \(b_n=\frac{n}{n+1}\), is the ratio \(\frac{b_{n+1}}{b_n}\) less than or greater than \(1\)? Show the exact calculation.
- Does proving that \(b_{n+1}\le b_n\) eventually, by itself, establish the hypotheses of the Alternating Series Test? What else is required?
