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Ratio, Root, and Test Selection · Tutorial 684 of 998

Binomial-Coefficient Series and the Ratio Test

Learn to simplify binomial-coefficient ratios before applying the Ratio Test, including for central coefficients and exponential weights.

Advanced 9 min read

What You'll Learn

  • Define binomial coefficients and derive ratios for fixed lower indices.
  • Derive the successive-term ratio for central binomial coefficients.
  • Find the limiting growth factor for coefficients of the form binomial pn choose qn.
  • Apply the Ratio Test to exponentially weighted binomial-coefficient series.
  • Identify boundary cases and correct common ratio and limit errors.

Why Binomial Coefficients Need Careful Ratios

Binomial coefficients can grow rapidly as their indices increase. When they occur in a series, an exponential weight may either outweigh that growth or be outweighed by it. The Ratio Test makes this comparison precise: the key is to compute the ratio of consecutive terms accurately and simplify it before taking a limit.

The indices matter. A binomial coefficient with a fixed lower index, such as \(\binom{n}{3}\), grows in a different way from a central coefficient such as \(\binom{2n}{n}\). Their successive-term ratios have different limits, so they lead to different convergence thresholds when multiplied by an exponential factor.

Definition: For integers \(n\ge 0\) and \(0\le k\le n\), the binomial coefficient is $$ \binom{n}{k}=\frac{n!}{k!(n-k)!}. $$ Here \(n!=1\cdot 2\cdots n\) for \(n\ge1\), and \(0!=1\).

The Ratio Test for Exponentially Weighted Terms

For a series \(\sum a_n\), apply the Ratio Test by finding the limit \(L=\lim_{n\to\infty}|a_{n+1}/a_n|\), provided the ratios are defined for all sufficiently large \(n\). If \(L<1\), the series converges absolutely. If \(L>1\), its terms do not tend to zero, so the series diverges. If \(L=1\), the test gives no conclusion.

When a term contains \(r^n\), the exponential factor contributes \(|r|\) to the ratio of consecutive absolute values. The remaining factor comes from the binomial coefficients. Thus the limit compares the size of the exponential weight with the binomial coefficient's successive-term growth.

Fixed Lower Index: \(\binom{n}{k}\)

First keep \(k\) fixed while \(n\) increases. From the factorial definition, for \(n\ge k\),

$$ \binom{n}{k}=\frac{n!}{k!(n-k)!}. $$

For \(k\ge1\), the ratio of consecutive coefficients is

$$ \frac{\binom{n+1}{k}}{\binom{n}{k}} = \frac{(n+1)!}{k!(n+1-k)!} \cdot \frac{k!(n-k)!}{n!} = \frac{n+1}{n+1-k}. $$

For fixed \(k\), this ratio tends to \(1\). Consequently, multiplying \(\binom{n}{k}\) by \(r^n\) gives a ratio limit of \(|r|\), as long as \(r\ne0\). The exponential weight determines the Ratio Test threshold in this case.

Worked Example: A Fixed Lower Index with an Exponential Weight

Consider \(\sum_{n=2}^{\infty}\binom{n}{2}(3/8)^n\). Write its positive term as \(a_n=\binom{n}{2}(3/8)^n\). For \(n\ge2\), its consecutive-term ratio is

$$ \frac{a_{n+1}}{a_n} = \frac{\binom{n+1}{2}}{\binom{n}{2}}\cdot\frac{3}{8} = \frac{(n+1)n/2}{n(n-1)/2}\cdot\frac{3}{8} = \frac{3(n+1)}{8(n-1)}. $$

Taking the limit gives

$$ L=\lim_{n\to\infty}\frac{3(n+1)}{8(n-1)} =\frac{3}{8}<1. $$

The terms are nonnegative, and the Ratio Test therefore proves that the series converges absolutely. As a check on the simplification, the coefficient ratio alone is \((n+1)/(n-1)\), which tends to \(1\); including the weight multiplies that ratio by \(3/8\), as required.

Central Binomial Coefficients

The central binomial coefficient in row \(2n\) is \(\binom{2n}{n}\). Its two indices both change when \(n\) increases, so the fixed-\(k\) ratio does not apply. Start from the factorial definition and compare consecutive central coefficients directly:

$$ \frac{\binom{2n+2}{n+1}}{\binom{2n}{n}} = \frac{(2n+2)!}{(n+1)!(n+1)!} \cdot \frac{n!n!}{(2n)!} = \frac{(2n+2)(2n+1)}{(n+1)^2} = \frac{2(2n+1)}{n+1}. $$

The exact ratio tends to \(4\), not \(1\). Thus central binomial coefficients have a successive-term growth factor of \(4\). With an exponential weight \(x^n\), the absolute ratio has limit \(4|x|\).

Result: For \(\sum_{n=0}^{\infty}\binom{2n}{n}x^n\), the Ratio Test proves absolute convergence when \(4|x|<1\), or \(|x|<1/4\). It proves divergence when \(4|x|>1\), or \(|x|>1/4\). At \(|x|=1/4\), the ratio limit is \(1\), so this test alone is inconclusive.

Worked Example: A Weighted Central Binomial Series

Consider \(\sum_{n=0}^{\infty}\binom{2n}{n}(1/5)^n\). For \(n\ge0\), let \(a_n=\binom{2n}{n}(1/5)^n\). Then

$$ \frac{a_{n+1}}{a_n} = \frac{\binom{2n+2}{n+1}}{\binom{2n}{n}}\cdot\frac{1}{5} = \frac{2(2n+1)}{5(n+1)}. $$

Taking the limit,

$$ L=\lim_{n\to\infty}\frac{2(2n+1)}{5(n+1)} =\frac{4}{5}<1. $$

Therefore the series converges absolutely. The calculation checks against the general threshold: \(4|x|=4/5\) when \(x=1/5\).

A Wider Family: \(\binom{pn}{qn}\)

The same method handles many binomial coefficients whose two indices grow proportionally. Let \(p\) and \(q\) be fixed positive integers with \(1\le q<p\), and put \(B_n=\binom{pn}{qn}\). The factorial definition gives

$$ \frac{B_{n+1}}{B_n} = \frac{\prod_{j=1}^{p}(pn+j)} {\left(\prod_{j=1}^{q}(qn+j)\right) \left(\prod_{j=1}^{p-q}((p-q)n+j)\right)}. $$

There are \(p\) factors in the numerator and \(q+(p-q)=p\) in the denominator. Factor \(n\) from every factor. The powers of \(n\) cancel, leaving

$$ \frac{B_{n+1}}{B_n} = \frac{\prod_{j=1}^{p}(p+j/n)} {\left(\prod_{j=1}^{q}(q+j/n)\right) \left(\prod_{j=1}^{p-q}(p-q+j/n)\right)}. $$

As \(n\) tends to infinity, each \(j/n\) tends to zero. Since these are finite products, the ratio therefore tends to

$$ \lim_{n\to\infty}\frac{B_{n+1}}{B_n} = \frac{p^p}{q^q(p-q)^{p-q}}. $$

For a series with terms \(B_nx^n\), the absolute ratio limit is consequently \(|x|p^p/[q^q(p-q)^{p-q}]\). Values below \(1\) give absolute convergence by the Ratio Test, and values above \(1\) give divergence. A value equal to \(1\) remains inconclusive by this test.

Worked Example: A Series with \(\binom{3n}{n}\)

Consider \(\sum_{n=1}^{\infty}\binom{3n}{n}(1/10)^n\). Set \(B_n=\binom{3n}{n}\). Directly from the factorial definition,

$$ \frac{B_{n+1}}{B_n} = \frac{(3n+3)(3n+2)(3n+1)} {(n+1)(2n+2)(2n+1)}. $$

Since \(3n+3=3(n+1)\) and \(2n+2=2(n+1)\), cancel those common index factors:

$$ \frac{B_{n+1}}{B_n} = \frac{3(3n+2)(3n+1)}{2(n+1)(2n+1)}. $$

Including the exponential weight gives

$$ \frac{a_{n+1}}{a_n} = \frac{3(3n+2)(3n+1)}{20(n+1)(2n+1)} \longrightarrow \frac{3\cdot3\cdot3}{20\cdot1\cdot2} = \frac{27}{40}<1. $$

The series converges absolutely. The general formula gives the same check: \(p^p/[q^q(p-q)^{p-q}]=3^3/(1^1 2^2)=27/4\), and multiplying by \(1/10\) gives \(27/40\).

Reciprocal Central Coefficients and a Different Threshold

Putting a central binomial coefficient in the denominator reverses its contribution to the ratio. For \(a_n=x^n/\binom{2n}{n}\), with \(x\ne0\),

$$ \left|\frac{a_{n+1}}{a_n}\right| = |x|\frac{\binom{2n}{n}}{\binom{2n+2}{n+1}} = |x|\frac{n+1}{2(2n+1)} \longrightarrow \frac{|x|}{4}. $$

Thus the Ratio Test proves absolute convergence when \(|x|<4\) and divergence when \(|x|>4\). At \(|x|=4\), the ratio test has limit \(1\). In this particular family, the series also diverges at the boundary: for \(x=4\) or \(x=-4\), the absolute value of its \(n\)-th term is \(4^n/\binom{2n}{n}\ge1\), because \(\binom{2n}{n}\le\sum_{j=0}^{2n}\binom{2n}{j}=4^n\). The terms therefore do not tend to zero. This extra argument is needed; the Ratio Test itself does not settle the boundary.

Common Ratio Errors

A frequent mistake is to treat the central coefficient as though it had a fixed lower index, or to cancel only part of the factorial ratio. For example, an incorrect simplification is

$$ \frac{\binom{2n+2}{n+1}}{\binom{2n}{n}} \ \text{incorrectly simplified to}\ \frac{2n+1}{n+1} \longrightarrow 2. $$

This drops the factor \(2\) contributed by \(2n+2=2(n+1)\). The correct calculation retains it:

$$ \frac{(2n+2)(2n+1)}{(n+1)^2} = \frac{2(2n+1)}{n+1} \longrightarrow 4. $$

That missing factor changes the convergence threshold by a factor of two. A second common error is to treat a ratio limit of \(1\) as evidence of convergence or divergence. It is neither: the Ratio Test is inconclusive in that case, and a different argument is required if the boundary must be classified.

A Reliable Procedure

For a binomial-coefficient series with an exponential weight, use the following sequence of steps. Keep the indices visible until the factorial ratio has been simplified; only then take the limit.

1
Identify how the indices vary.
Check whether the lower index is fixed, whether the coefficient is central, or whether both indices grow proportionally.
2
Form the consecutive-term ratio.
Include the exponential factor: \(r^n\) contributes \(|r|\) to the absolute ratio.
3
Simplify exactly.
Use factorial cancellation before evaluating the limit, and verify that no index-dependent factor has been lost.
4
Apply the Ratio Test carefully.
A limit below \(1\) gives absolute convergence; a limit above \(1\) gives divergence; a limit of \(1\) requires another test or argument.

These calculations matter because binomial coefficients appear in combinatorial sums and in series whose terms combine rapid coefficient growth with exponential decay. The exact ratio identifies the competition without needing to estimate each term separately. The essential discipline is to distinguish fixed-index coefficients from central or proportionally indexed coefficients, and to treat the Ratio Test's boundary honestly.

Check Your Understanding

Use the factorial definition and the Ratio Test where appropriate. State clearly when the test is inconclusive.

  1. Find \(\lim_{n\to\infty}\binom{n+1}{4}/\binom{n}{4}\). What does this imply for the ratio limit of \(\binom{n}{4}(2/3)^n\)?
  2. Derive the exact ratio \(\binom{2n+2}{n+1}/\binom{2n}{n}\) and give its limit.
  3. For which values of \(|x|\) does the Ratio Test prove convergence of \(\sum_{n=0}^{\infty}\binom{2n}{n}x^n\)? For which values does it prove divergence?
  4. Find the ratio limit for terms \(\binom{4n}{n}(1/20)^n\), and classify the series using the Ratio Test.
  5. What does a Ratio Test limit of \(1\) establish, and what does it fail to establish?

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