Why Small Algebra Errors Matter
A ratio or root test can reduce a complicated series to a short limit. The conclusion is only as reliable as the algebra used to reach that limit. In particular, factorials shift by several factors at once, cancellation applies only to matching factors, and an exponent outside parentheses applies to the entire quantity inside them.
These errors can be difficult to notice because an incorrect expression may still look plausible. A reliable calculation keeps the original term visible, records each index change explicitly, and checks the simplified expression against a direct calculation for at least one index.
The Test Conditions to Keep in View
For a series \(\sum_{n=1}^{\infty} a_n\), the ratio test examines \(\left|a_{n+1}/a_n\right|\), provided the terms are nonzero from some point onward. If this ratio has limit \(L<1\), the series converges absolutely. If \(L>1\), the terms do not tend to zero and the series diverges. If \(L=1\), the test gives no conclusion.
The root test examines \(\sqrt[n]{|a_n|}\). If this quantity has limit \(L<1\), the series converges absolutely; if \(L>1\), it diverges; and if \(L=1\), the test is inconclusive. The calculations in this tutorial focus on getting these quantities right, not on choosing a different test when a limit equals one.
Factorials: Shift the Index, Then Expand
The factorial \(m!\), for a nonnegative integer \(m\), is the product \(1\cdot2\cdots m\), with \(0!=1\). A shift in its argument usually introduces more than one factor. For instance, advancing \(2n\) to \(2n+2\) adds two factors, not one.
This follows directly from the factorial definition: \((m+r)!=(m+r)(m+r-1)\cdots(m+1)m!\). Dividing both sides by \(m!\) leaves precisely the \(r\) new factors. This identity is a useful check whenever a ratio changes the index in a factorial.
Worked Example: A Two-Factor Shift
Consider the positive-term series \(\sum_{n=1}^{\infty} a_n\), where \(a_n=n!/(2n)!\). To apply the ratio test, write both terms in the quotient before canceling:
Now use \((n+1)!/n!=n+1\) and \((2n+2)!/(2n)!=(2n+2)(2n+1)\):
Therefore, \(\lim_{n\to\infty}a_{n+1}/a_n=0<1\), so the series converges by the ratio test. Check the algebra at \(n=2\): \(a_2=2!/4!=1/12\), while \(a_3=3!/6!=1/120\). Thus \(a_3/a_2=(1/120)/(1/12)=1/10\). The formula gives \(1/[2(2\cdot2+1)]=1/10\), as required.
A common index-shift error is to replace \((2n+2)!/(2n)!\) by only \(2n+2\). That omits the factor \(2n+1\). It would give the incorrect ratio \((n+1)/(2n+2)=1/2\), rather than \(1/[2(2n+1)]\). Both limits happen to be below one here, but the expressions and limits are not interchangeable; in another problem, the error could change the test's conclusion.
Cancel Factors, Not Similar-Looking Expressions
Cancellation is division of a numerator and denominator by the same nonzero factor. It does not allow a whole expression to disappear just because it resembles a factor elsewhere. In particular, \(3n+3\) is not \(n+1\): it equals \(3(n+1)\). If the numerator contains \((n+1)^2\), canceling one copy of \(n+1\) leaves another copy, as well as the factor \(3\) in the denominator.
Worked Example: Canceling One Matching Factor
Let \(a_n=(n!)^2/(3n)!\). The ratio is
Since \(3n+3=3(n+1)\), cancel one matching factor \(n+1\):
The numerator has degree one and the denominator has degree two, so this ratio tends to zero. More explicitly, dividing numerator and denominator by \(n^2\) shows the limit is \(0\). The ratio test gives convergence. Check at \(n=1\): \(a_1=1/6\), \(a_2=4/720=1/180\), and \(a_2/a_1=1/30\). The simplified formula gives \(2/[3\cdot5\cdot4]=1/30\), confirming the result.
The incorrect cancellation would be to erase \((n+1)^2\) against \(3(n+1)\) and leave only \(1/[(3n+2)(3n+1)]\). That loses both the remaining \(n+1\) and the factor \(3\). The correct quotient retains them: \((n+1)/[3(3n+2)(3n+1)]\).
Take the Root of the Entire Term
When using the root test, the exponent \(1/n\) applies to the complete absolute value of the term. Products, quotients, and powers must all remain inside that operation until they have been handled correctly. For positive \(x\) and \(y\), \(\sqrt[n]{x/y}=\sqrt[n]{x}/\sqrt[n]{y}\), and \(\sqrt[n]{x^k}=x^{k/n}\). In particular, \(\sqrt[n]{c^n}=c\) for \(c>0\), not \(c^n\).
Worked Example: Preserving the Exponential Denominator
Consider the positive terms
The nth root must be applied to both the numerator and denominator:
Using the standard limit \(\lim_{n\to\infty}(1+1/n)^n=e\), the root limit is \(e/3<1\). Therefore, \(\sum_{n=1}^{\infty}a_n\) converges by the root test. To check the exponent handling at \(n=2\), the term is \(a_2=(3/2)^4/9=9/16\). Its square root is \(3/4\), and the root formula gives \((3/2)^2/3=3/4\) as well.
A concrete wrong calculation is to take the root of the numerator but forget the denominator, obtaining the limit \(e\) and claiming divergence. The denominator contributes \((3^n)^{1/n}=3\), so the actual limit is \(e/3\), which is less than one. Another error is to write \((3^n)^{1/n}=3^n\); the exponent \(n\) is divided by \(n\), leaving \(3^1=3\).
A Reliable Audit for a Test Calculation
A brief sequence of checks can catch most index and exponent errors before they affect the conclusion.
Keep all factorial arguments, parentheses, and powers visible before forming a ratio or root.
For each factorial, list every factor introduced by the change in its argument.
Factor expressions such as \(3n+3=3(n+1)\) before canceling, and retain any unmatched factors.
For a quotient, take roots of numerator and denominator; divide every exponent by \(n\).
Compute one exact quotient or root directly and compare it with the simplified formula.
The small-index check does not prove that an algebraic identity is valid for every \(n\). It is a diagnostic: if the two calculations disagree even once, the simplification is wrong. The general justification still comes from expanding factorials and applying valid algebraic rules.
When a Correct Limit Is Not a Conclusion
Careful algebra and correct use of the test are separate requirements. A perfectly simplified ratio or root limit equal to one is inconclusive; it does not establish convergence or divergence. Conversely, a limit below one proves absolute convergence under the stated test conditions, while a limit above one proves divergence. Do not let a confident-looking simplification hide either a calculation error or an inconclusive test result.
Check Your Understanding
For each question, identify the index or exponent operation that must be checked before applying a convergence test.
- Expand \((3n+2)!/(3n)!\) as a product of consecutive factors.
- Simplify \((n+1)^2/[(4n+4)(4n+3)]\) by factoring the first denominator factor and canceling only what matches.
- Find the nth root of \((2+1/n)^{n^2}/7^n\), leaving the result in a form suitable for taking a limit.
- A ratio-test calculation gives a limit of \(1\). What conclusion does the ratio test provide?
- For any ratio or root calculation, name one direct small-index check that could reveal an algebra error.
