Introduction
At the boundary of a power series, the terms can become much more informative than the radius of convergence alone. In this tutorial, the key coefficient is the normalized central binomial coefficient \(c_n=\binom{2n}{n}/4^n\). It tends to zero, but only at the rate of a reciprocal square root. That rate is slow enough to make the positive boundary series diverge, while an alternating sign at the other endpoint produces convergence.
We study the power series
At \(x=4\), the factor \((x/4)^n\) is \(1\); at \(x=-4\), it is \((-1)^n\). Thus the endpoint classifications depend on the same positive coefficients, with alternating signs appearing only at the left endpoint.
The Asymptotic Size of the Coefficients
The decisive estimate is
Here \(a_n\sim b_n\) means that \(a_n/b_n\to 1\). In particular, the estimate says not merely that \(c_n\to0\), but that its size is comparable to \(1/\sqrt{n}\). We derive the estimate using integrals of powers of sine.
Since \(I_0=\pi/2\) and \(I_1=1\), repeated use of this recurrence gives
The product identities follow by writing the even factors as \(2\cdot4\cdots2n=2^n n!\) and the odd factors as \((2n)!/(2^n n!)\). Also, \(I_{m+1}\le I_m\), because \(0\le\sin t\le1\) on the interval of integration. Applying the recurrence once more gives \(I_{m+2}/I_m=(m+1)/(m+2)\). Therefore
Both bounds tend to \(1\), so \(I_{m+1}/I_m\to1\). In particular, this holds when \(m=2n\). Using the formulas for \(I_{2n}\) and \(I_{2n+1}\),
It follows that \(\pi(2n+1)c_n^2\to2\), and hence \(n c_n^2\to1/\pi\). Since every \(c_n\) is positive, taking square roots proves \(\sqrt n\,c_n\to1/\sqrt{\pi}\), which is the stated asymptotic estimate.
Worked Example: Checking the First Coefficients
Worked Example: Checking the First Coefficients
Compute \(c_0,c_1,c_2,c_3\), and verify that they agree with the coefficient recurrence.
From the definition,
For \(n\ge0\), the factorial definition also gives
For \(n=0\), this gives \(c_1=c_0(1/2)=1/2\). For \(n=1\), it gives \(c_2=c_1(3/4)=(1/2)(3/4)=3/8\). For \(n=2\), it gives \(c_3=c_2(5/6)=(3/8)(5/6)=15/48=5/16\). Each value agrees with the direct calculation, checking both the coefficients and the recurrence.
The Right Endpoint: Divergence at \(x=4\)
At \(x=4\), substituting into the series gives
The terms tend to zero, so the term test alone does not decide convergence. The asymptotic estimate gives the stronger information needed. For all sufficiently large \(n\), it implies, for example, \(c_n\ge 1/(2\sqrt{\pi n})\). The series \(\sum_{n=1}^{\infty}1/\sqrt n\) is a divergent \(p\)-series, with \(p=1/2\). Direct comparison therefore shows that \(\sum c_n\) diverges. Since its terms are positive, its partial sums grow without bound.
Worked Example: Classifying the Series at \(x=4\)
Consider the endpoint series from \(n=1\):
Its \(n\)th term is \(c_n\), and the derived limit is \(\sqrt n\,c_n\to1/\sqrt\pi\). Because \(1/\sqrt\pi>0\), there is an integer \(N\) such that for every \(n\ge N\),
Consequently, the tail of the series is at least the corresponding tail of a positive constant times the divergent \(p\)-series:
As \(M\to\infty\), the right-hand partial sums are unbounded. The left-hand partial sums are therefore unbounded as well, so the endpoint series diverges. The omitted term \(c_0=1\) is only a finite initial term and cannot change this verdict. This comparison verifies the divergence even though \(c_n\to0\).
The Left Endpoint: Conditional Convergence at \(x=-4\)
At \(x=-4\), substitution gives the alternating series
Each \(c_n\) is positive. The recurrence \(c_{n+1}/c_n=(2n+1)/(2n+2)\) shows that \(c_{n+1}<c_n\), and the asymptotic estimate shows that \(c_n\to0\). Thus the alternating-series test applies: a series with terms \((-1)^n b_n\) converges when \(b_n\) is nonincreasing, nonnegative, and tends to zero.
This convergence is not absolute. Taking absolute values removes the alternating signs and leaves \(\sum c_n\), which diverges by the comparison at \(x=4\). The series at \(x=-4\) is therefore conditionally convergent: it converges, but the series of absolute values does not.
Worked Example: Classifying the Series at \(x=-4\)
The first terms are obtained from the coefficients already computed:
To verify convergence using partial sums, define \(S_N=\sum_{n=0}^{N}(-1)^n c_n\). Since \(c_n\) is strictly decreasing, for every \(m\ge0\),
Thus the even partial sums decrease, while the odd partial sums increase. Each odd partial sum lies below the even partial sum immediately after it, since \(S_{2m+2}-S_{2m+1}=c_{2m+2}>0\). Equivalently, the gap between the even partial sum \(S_{2m}\) and the odd partial sum immediately after it is
The odd partial sums are bounded above by the even partial sums, and the even partial sums are bounded below by the odd ones. Both subsequences therefore converge; because their gap tends to zero, they have the same limit. Hence the full sequence of partial sums converges. For a numerical check, \(S_0=1\), \(S_1=1/2\), \(S_2=7/8\), and \(S_3=9/16\); these values satisfy the stated ordering of odd and even partial sums. Finally, the absolute-value series is exactly \(\sum c_n\), already shown to diverge. This confirms conditional, not absolute, convergence.
A Common Pitfall: “The Terms Go to Zero, So the Series Converges”
A tempting but incorrect argument at \(x=4\) is: “Since \(c_n\to0\), the series \(\sum c_n\) converges.” The calculation only establishes that the individual terms become small. It does not establish that their accumulated sum stays finite. Here the specific rate matters: \(c_n\) behaves like \(1/\sqrt{\pi n}\), and the sum of terms of that size diverges.
At \(x=-4\), the conclusion is different for a specific reason: the alternating signs permit cancellation, and the positive term sizes decrease to zero. That proves convergence of the alternating series, but it does not prove convergence after taking absolute values. The absolute-value series returns to the divergent positive series.
Check Your Understanding
Use the coefficient asymptotic and the endpoint tests to answer these questions.
- What does \(c_n\sim1/\sqrt{\pi n}\) mean in terms of a limit?
- Why does \(c_n\to0\) fail to prove convergence of the series at \(x=4\)?
- Which two conditions on \(c_n\) allow the alternating-series test at \(x=-4\) to apply?
- What series results from taking the absolute values of the terms at \(x=-4\), and what is its convergence classification?
- Why do the odd and even partial sums in the alternating example have the same limit?
