Why the Starting Index Matters
A series may begin at \(n=1\), but it does not have to. For example, a formula involving \(\ln n\) may be used only for \(n\ge 2\), and a series whose first term is indexed by \(n=5\) need not be defined at all for smaller inputs. The integral test still applies when its conditions hold from the series’ actual starting index onward.
The key is to keep two questions separate. First, is the defining function valid and suitably decreasing where the series terms begin? Second, does the corresponding improper integral converge? A function’s failure to be defined before the first included term does not by itself prevent the test from applying.
The Integral Test at a Shifted Starting Index
For a series \(\sum_{n=m}^{\infty} a_n\), the starting index \(m\) is the first index included. The integral test requires a function \(f\) that agrees with the terms, \(f(n)=a_n\), and is continuous, positive, and decreasing on \([m,\infty)\). “Decreasing” here permits nonincreasing values; in the examples below, the functions are strictly decreasing.
The lower limit is \(m\), not automatically \(1\). If the series begins later than \(m\), changing the lower limit by a finite amount does not change whether the improper integral converges, provided the function is continuous on the finite interval being added or removed. If the function is not defined there, do not include that interval: start the integral where the function and the test’s conditions are valid.
Why Finite Initial Terms Do Not Change Convergence
A series converges when its sequence of partial sums has a finite limit. Omitting or adding finitely many terms changes each later partial sum by a fixed finite amount, so it cannot change whether those partial sums have a finite limit. This is why a test can be applied to a tail, even if its conditions hold only from some index onward.
Choose an index \(N\) after which the terms and the defining function satisfy the integral-test conditions.
Apply the integral test to \(\sum_{n=N}^{\infty}a_n\), using \(\int_N^\infty f(x)\,dx\).
The original series differs from that tail by only finitely many terms, so it has the same convergence behavior.
For completeness, if the original series begins at \(m<N\), then for every \(K\ge N\), its partial sum through \(K\) is the sum of its first \(N-m\) terms plus the tail’s partial sum through \(K\). The first quantity is a fixed finite number. Adding it preserves the existence or nonexistence of a finite limit. Thus testing the tail decides convergence of the full series.
Worked Example: A Logarithmic Series Starting at \(n=4\)
Worked Example: A Logarithmic Series Starting at \(n=4\)
Determine whether \(\sum_{n=4}^{\infty}\frac{1}{n(\ln n)^2}\) converges. Set \(f(x)=\frac{1}{x(\ln x)^2}\). Its domain is \(x>1\), so it is defined on the full interval \([4,\infty)\). For \(x\ge4\), \(f(x)>0\), and
because \(x>1\) implies \(\ln x>0\). Thus \(f\) is continuous, positive, and decreasing where required. Evaluate the integral from the actual starting index:
As \(b\to\infty\), \(\ln b\to\infty\), so \(\frac{1}{\ln b}\to0\). Therefore
The integral test shows that the series converges. The function is also defined on \((1,\infty)\), but there is no need to integrate from \(1\), where this improper integral would not be finite. Starting at \(4\) is valid and directly matches the series.
Worked Example: A Series Starting Where the Function Becomes Valid
Worked Example: A Series Starting Where the Function Becomes Valid
Determine whether \(\sum_{n=2}^{\infty}\frac{1}{n\ln n}\) converges. The expression \(\frac{1}{x\ln x}\) is undefined at \(x=1\), but the series begins at \(n=2\). Define \(f(x)=\frac{1}{x\ln x}\) for \(x>1\). It is continuous and positive on \([2,\infty)\), and
All integral-test conditions therefore hold from the first included index. For \(b>2\),
As \(b\to\infty\), \(\ln(\ln b)\to\infty\). Hence \(\int_2^\infty\frac{dx}{x\ln x}\) diverges, and the integral test shows that the series diverges. The value at \(x=1\) is irrelevant: neither the series nor the integral used in the test includes that input.
Reindexing Does Not Change the Series
Sometimes a series is written using an index that begins at \(1\), even though the formula involves larger inputs. Reindexing makes the relationship explicit. If \(n=k+c\), where \(c\) is a fixed integer, then \(k=1,2,\ldots\) corresponds exactly to \(n=c+1,c+2,\ldots\). Each term is retained once, in the same order; only its label changes.
Worked Example: Reindexing a Series Before Applying the Test
Worked Example: Reindexing a Series Before Applying the Test
Determine whether \(\sum_{k=1}^{\infty}\frac{1}{(k+3)(\ln(k+3))^2}\) converges. Set \(n=k+3\). When \(k=1\), \(n=4\), and as \(k\) runs through the positive integers, \(n\) runs through \(4,5,6,\ldots\). Thus the series is exactly
For \(f(x)=\frac{1}{x(\ln x)^2}\), the domain is \(x>1\), and for \(x\ge4\) the function is positive and continuous. Its derivative is
So the integral test applies. Using the correct lower limit,
The integral is finite, so the reindexed series converges. As a check on the reindexing, the first term in either expression is \(\frac{1}{4(\ln4)^2}\): setting \(k=1\) in the original expression and \(n=4\) in the reindexed expression gives the same term.
A Common Mistake: Forcing the Integral to Start at 1
For the series \(\sum_{n=2}^{\infty}\frac{1}{n\ln n}\), a tempting but invalid calculation is
This does not evaluate the integral: the integrand is not defined at \(x=1\), and the displayed antiderivative cannot be substituted there. More importantly, the series has no term with \(n=1\), so this is not the integral the test calls for. The correction is to use \(\int_2^\infty\frac{dx}{x\ln x}\). That integral diverges, as shown above, so the conclusion is still divergence—but it must follow from a valid integral.
A related indexing error is to change \(n=k+3\) in the index while leaving the formula as \(a_k\). That would generally produce different terms. Check the first included term after every reindexing: matching first terms is a quick way to catch an incorrect shift.
Why the Distinction Is Useful
Many familiar functions have domain restrictions. Logarithms require positive inputs, and expressions such as \(\ln(\ln x)\) impose further restrictions. A series may nonetheless be perfectly well-defined because it starts at an index large enough to meet those restrictions. The integral test concerns the function on the interval where the series starts and continues, not at every smaller real number.
At the same time, finite-term flexibility is not permission to ignore a problem at an included index. If a term in the stated series is undefined, the series as written is not defined. If only the function fails to satisfy the test’s conditions for a few initial inputs, one may instead apply the test to a valid tail—but only if the series terms themselves are defined. Keeping these cases distinct prevents both unnecessary concern about unused inputs and careless use of the test.
Check Your Understanding
For each question, identify the series’ actual starting input before deciding which integral to use.
- For \(\sum_{n=3}^{\infty}\frac{1}{n(\ln n)^2}\), what is the appropriate lower limit for the integral test?
- Why does the fact that \(\frac{1}{x\ln x}\) is undefined at \(x=1\) not prevent testing \(\sum_{n=2}^{\infty}\frac{1}{n\ln n}\)?
- When \(n=k+4\) and \(k\) starts at \(1\), what is the first value of \(n\)?
- If the integral-test conditions hold only for \(n\ge N\), what must be checked before applying the test to that tail?
- Explain why adding or removing finitely many defined terms cannot change convergence.
