Why Check the Function Before Integrating?
A series term may be written as \(a_n=f(n)\), connecting a sequence of values to a function of a real variable. Before using the Integral Test on a tail of the series, it is necessary to verify the relevant properties of that function: it must be positive, continuous, and decreasing throughout the real interval being considered.
These conditions do not have to hold from the very first term. If they hold from some integer \(N\) onward, the Integral Test can be applied to the tail beginning at \(N\). The terms before \(N\) are finite in number, so they do not determine whether the series converges. The key practical task is therefore to identify a suitable starting point and justify the conditions on the entire interval beyond it.
What Must Be Verified
For a series with terms \(a_n=f(n)\), the Integral Test applies to a tail when \(f\) is positive, continuous, and decreasing on \([N,\infty)\) for some integer \(N\), and the corresponding improper integral is considered. Here the emphasis is on verifying those function conditions, not on evaluating the integral.
Each condition is about the whole real interval \([N,\infty)\), not just the integer inputs. Checking a few values of \(f(n)\) can show what happens at those particular terms, but it cannot establish that \(f\) is decreasing between them. Derivative signs provide a useful sufficient condition for monotonicity.
The derivative criterion addresses decrease only. Positivity and continuity still need their own checks. For example, a negative derivative does not make a function positive, and a function with a favorable derivative sign may still be undefined at a point in the interval.
Why a Derivative Sign Proves Decrease
The justification uses the Mean Value Theorem. It is enough to compare any two points in the interval, rather than checking every value separately.
Let \(x\) and \(y\) be any real numbers with \(N\le x<y\).
Continuity on \([x,y]\) and differentiability on \((x,y)\) give a point \(c\in(x,y)\) such that \(f(y)-f(x)=f'(c)(y-x)\).
Since \(f'(c)\le 0\) and \(y-x>0\), their product is nonpositive. Thus \(f(y)-f(x)\le 0\), so \(f(y)\le f(x)\).
Because the two points were arbitrary, the inequality holds for every \(N\le x<y\). That is precisely the definition of nonincreasing. If \(f'(c)<0\) for every possible \(c\), then the same argument gives \(f(y)<f(x)\), proving strict decrease.
Worked Example: Decrease Begins After the First Terms
Worked Example: Finding a Tail Where the Derivative Is Negative
Consider \(a_n=f(n)\), where \(f(x)=\dfrac{x+10}{x^2+100}\), for integers \(n\ge 1\). Find an integer \(N\) such that \(f\) is positive, continuous, and decreasing for \(x\ge N\).
First, \(x^2+100>0\) for every real \(x\), so \(f\) is continuous everywhere. For \(x\ge1\), both \(x+10>0\) and \(x^2+100>0\), hence \(f(x)>0\).
Differentiate using the quotient rule:
The denominator is positive. The derivative is negative when \(100-x^2-20x<0\), or equivalently \(x^2+20x-100>0\). For \(x\ge5\), \(x^2+20x\ge25+100=125>100\), so \(f'(x)<0\). Thus \(f\) is positive, continuous, and strictly decreasing on \([5,\infty)\), and \(N=5\) works.
The decrease does not begin at the first term. Indeed, \(f'(1)=79/10201>0\). The initial terms also illustrate the rise: \(f(1)=11/101\) and \(f(2)=12/104=3/26\). Since \(11\cdot26=286<303=3\cdot101\), \(f(1)<f(2)\). The derivative check establishes the needed behavior for the tail, not for every earlier term.
Check: At \(x=5\), the derivative formula gives \(f'(5)=(100-25-100)/125^2=-25/15625=-1/625<0\), consistent with the sign established for all \(x\ge5\).
Worked Example: Positivity Can Start Before Decrease
Worked Example: Choosing the First Integer Past a Threshold
Let \(f(x)=\dfrac{x-1}{x^2+1}\), and consider the terms \(a_n=f(n)\) for \(n\ge2\). Determine an integer \(N\) such that the function is positive, continuous, and decreasing for \(x\ge N\).
The denominator \(x^2+1\) is always positive, and the numerator \(x-1\) is positive for \(x>1\). Hence \(f(x)>0\) for \(x\ge2\). The denominator never vanishes, so \(f\) is continuous on \([2,\infty)\).
Differentiate:
The denominator is positive. The numerator is negative when \(x^2-2x-1>0\), which holds for \(x>1+\sqrt{2}\). Since \(\sqrt{2}<2\), we have \(1+\sqrt{2}<3\). Therefore \(f'(x)<0\) for every \(x\ge3\), so \(N=3\) works.
At \(x=2\), the derivative is \(f'(2)=1/25>0\), so the derivative does not have the required negative sign from the start. The terms at the boundary are \(f(2)=1/5\) and \(f(3)=2/10=1/5\). For the tail beginning at \(n=3\), the derivative criterion guarantees strict decrease from one real input to any larger one.
Check: Substituting \(x=3\) in the derivative formula gives \((-9+6+1)/(9+1)^2=-2/100=-1/50<0\), as required.
Worked Example: Check the Domain of the Tail
Worked Example: A Function with an Earlier Singularity
Consider \(a_n=\dfrac{1}{n^2-1}\) for integers \(n\ge2\), represented by \(f(x)=\dfrac{1}{x^2-1}\). Verify the function conditions on a suitable interval.
The denominator factors as \((x-1)(x+1)\). It vanishes at \(x=1\) and \(x=-1\), but neither point belongs to \([2,\infty)\). For every \(x\ge2\), \(x^2-1\ge3>0\), so \(f(x)>0\), and the rational function is continuous on \([2,\infty)\).
Its derivative is
For \(x\ge2\), the numerator \(-2x\) is negative and the denominator is positive. Thus \(f'(x)<0\) throughout the interval, proving strict decrease. The integer starting point \(N=2\) satisfies all three conditions.
Check: The first two terms are \(a_2=1/(4-1)=1/3\) and \(a_3=1/(9-1)=1/8\), with \(1/3>1/8\), in agreement with the derivative conclusion. The singularity at \(x=1\) does not obstruct using the tail interval \([2,\infty)\); it would obstruct claiming continuity on an interval that includes \(1\).
Common Pitfall: An Increasing Denominator Is Not Enough
A quotient can increase even while its denominator increases. Looking only at a denominator’s behavior does not determine the derivative sign of the whole function. For instance, in the first worked example both \(x+10\) and \(x^2+100\) increase for \(x\ge1\), but the quotient initially increases as well.
A concrete incorrect argument would be: “Since \(x^2+100\) increases, \((x+10)/(x^2+100)\) decreases for every \(x\ge1\).” This conclusion is false. The derivative calculation gives
so the function is increasing near \(x=1\), not decreasing there. The correction is to differentiate the entire quotient and analyze the sign of the resulting numerator, while separately checking the denominator and the function’s domain.
Choosing the Starting Index Carefully
Derivative sign conditions are often first obtained for real \(x\), while a series tail begins at an integer index. If a calculation shows \(f'(x)<0\) whenever \(x>\alpha\), choose an integer \(N\) such that every \(x\ge N\) lies in that range. When the inequality is strict only for \(x>\alpha\), choosing an integer strictly greater than \(\alpha\) avoids including a point where the derivative may be zero or where the sign condition has not been proved.
Then verify positivity and continuity on \([N,\infty)\) as well. The interval must be large enough for all three properties at once; a threshold that works for the derivative alone may not be enough if the function has a zero, a singularity, or a change of sign farther along the domain.
Check Your Understanding
For each question, identify the conditions that must be checked on the proposed tail, not merely at a few integer inputs.
- For \(f(x)=\dfrac{x+4}{x^2+25}\), what derivative-sign inequality would you solve to locate an interval of decrease?
- If \(f'(x)<0\) for every \(x>4\), which integer starting point ensures the derivative is negative throughout the tail interval?
- Why does continuity on \([N,\infty)\) need to be checked separately from the sign of \(f'(x)\)?
- If the required properties hold only for \(x\ge7\), what happens to the finitely many series terms before \(n=7\) when applying a test to the tail?
- Can checking \(f(5)>f(6)>f(7)\) prove that \(f\) is decreasing on \([5,\infty)\)? Explain what the derivative criterion adds.