Why Sums Preserve the Smaller Radius

Why the Smaller Radius Is Guaranteed

A power series can converge for some values of its variable and fail to converge for others. Its radius of convergence describes the size of the region around its center where convergence is guaranteed. When two power series have the same center, their terms can be added coefficient by coefficient. The key question is: how much of the convergence region must the resulting series retain?

The answer is a guarantee, not always an exact prediction. The sum converges inside the smaller of the two radii. It may converge farther still if coefficients cancel. This distinction is useful whenever a power series is formed by adding expressions whose individual convergence behavior is already known.

Definition: A power series centered at \(c\) is an expression of the form \(\sum_{n=0}^{\infty} a_n(x-c)^n\), where \(a_n\) are coefficients. Its radius of convergence \(R\) is the nonnegative number (possibly \(+\infty\)) such that the series converges absolutely when \(|x-c|<R\) and diverges when \(|x-c|>R\). Behavior at \(|x-c|=R\) must be tested separately.

Throughout, the two series must have the same center. Write them as

\sum_{n=0}^{\infty} a_n(x-c)^n \qquad\text{and}\qquad \sum_{n=0}^{\infty} b_n(x-c)^n.

Their coefficientwise sum is the power series \(\sum_{n=0}^{\infty}(a_n+b_n)(x-c)^n\). If the original radii are \(R_a\) and \(R_b\), the result has radius at least \(\min(R_a,R_b)\). “At least” matters: the sum need not have exactly that radius.

The Main Result and Its Proof

Theorem: Suppose two power series with the same center \(c\) have radii of convergence \(R_a\) and \(R_b\). Then the coefficientwise sum \(\sum_{n=0}^{\infty}(a_n+b_n)(x-c)^n\) converges absolutely for every \(x\) satisfying \(|x-c|<\min(R_a,R_b)\). Consequently, its radius of convergence is at least \(\min(R_a,R_b)\).

The reason is the triangle inequality. At a point strictly inside both radii, each original series converges absolutely. The absolute values of the combined terms are bounded by the sum of the absolute values of the original terms.

Choose a point inside the smaller radius.
Let \(|x-c|<\min(R_a,R_b)\). Then \(|x-c|<R_a\) and \(|x-c|<R_b\).

Use absolute convergence of each series.
By the convergence property of power series, both \(\sum |a_n||x-c|^n\) and \(\sum |b_n||x-c|^n\) are finite.

Bound the combined terms.
For every \(n\), the triangle inequality gives \(|a_n+b_n||x-c|^n\leq |a_n||x-c|^n+|b_n||x-c|^n\).

Conclude absolute convergence.
The sum of the right-hand side over \(n\) is finite, so comparison shows \(\sum |a_n+b_n||x-c|^n\) is finite as well.

This proves convergence at every point strictly inside the smaller radius. The radius of the sum therefore cannot be smaller than that radius. The argument does not make a claim about the boundary: at \(|x-c|=\min(R_a,R_b)\), one or both original series may converge or diverge, so that point requires separate analysis.

Calculating a Radius from Coefficients

For examples, the root test gives a convenient way to determine a power-series radius. If the coefficients are \(d_n\), then

R=\frac{1}{\limsup_{n\to\infty}|d_n|^{1/n}}.

Here \(1/0=+\infty\) and \(1/(+\infty)=0\). In particular, when \(|d_n|^{1/n}\) has a limit \(L>0\), the radius is \(1/L\). This formula identifies the radius, but it does not settle convergence at points on its boundary.

Worked Example: A Sum Whose Radius Is the Smaller One

Find the radius of the coefficientwise sum of

\sum_{n=0}^{\infty}\left(\frac{x}{2}\right)^n \qquad\text{and}\qquad \sum_{n=0}^{\infty}\left(\frac{x}{3}\right)^n.

The first series has coefficients \(a_n=2^{-n}\). Since \(|a_n|^{1/n}=1/2\) for \(n\geq1\), its radius is \(R_a=2\). The second has coefficients \(b_n=3^{-n}\), so \(|b_n|^{1/n}=1/3\) and \(R_b=3\). The theorem guarantees that their sum converges for \(|x|<2\).

Adding corresponding coefficients gives \(d_n=2^{-n}+3^{-n}\). For \(n\geq1\), factor out \(2^{-n}\):

|d_n|^{1/n} = \left(2^{-n}\left[1+\left(\frac{2}{3}\right)^n\right]\right)^{1/n} = \frac12\left[1+\left(\frac{2}{3}\right)^n\right]^{1/n} \longrightarrow \frac12.

Therefore the sum's radius is \(R=1/(1/2)=2\), exactly the smaller radius. As a numerical check inside the radius, take \(x=1\). The original geometric series have sums \(1/(1-1/2)=2\) and \(1/(1-1/3)=3/2\), for a total of \(7/2\). The coefficientwise sum gives the same value:

\sum_{n=0}^{\infty}\left(2^{-n}+3^{-n}\right) = \frac{1}{1-1/2}+\frac{1}{1-1/3} = 2+\frac32 = \frac72.

This check is valid because \(x=1\) lies inside both radii.

Cancellation Can Enlarge the Radius

The theorem gives a lower bound because it uses an inequality. That inequality may be strict: if \(a_n\) and \(b_n\) have opposite signs, adding them can make \(a_n+b_n\) much smaller than either coefficient. The original series may then have limited radii even though the combined coefficients decrease much faster.

Worked Example: Cancellation Increases the Radius from 1 to 2

Consider the two series centered at zero:

\sum_{n=0}^{\infty}x^n \qquad\text{and}\qquad \sum_{n=0}^{\infty}\left(-1+2^{-n}\right)x^n.

The first has coefficients \(a_n=1\), so \(\lim |a_n|^{1/n}=1\) and its radius is \(1\). The second has coefficients \(b_n=-1+2^{-n}\). Since \(b_n\to-1\), we have \(|b_n|^{1/n}\to1\), so its radius is also \(1\).

Adding the coefficients, rather than multiplying the series, gives

a_n+b_n=1+\left(-1+2^{-n}\right)=2^{-n}.

Thus the sum is \(\sum_{n=0}^{\infty}2^{-n}x^n\). Its coefficient root is \((2^{-n})^{1/n}=1/2\) for \(n\geq1\), so its radius is \(2\). This is larger than either original radius.

For a direct check, the \(N\)th partial sums of the two original series add as follows:

\sum_{n=0}^{N}x^n+ \sum_{n=0}^{N}\left(-1+2^{-n}\right)x^n = \sum_{n=0}^{N}2^{-n}x^n.

The right-hand side converges to \(1/(1-x/2)\) when \(|x|<2\). For example, at \(x=3/2\), the combined series converges to \(1/(1-3/4)=4\). The original series do not converge separately at that value, so this is a value of the combined power series, not a sum of two separately convergent series there.

This distinction is important. Where both original series converge, their sums can be added in the usual way, and the coefficientwise sum has that value. Outside their common convergence region, the combined series may still converge because of cancellation, but the two original series do not necessarily have individual sums there.

Worked Example: Cancellation with Sparse Coefficients

Let

A(x)=\sum_{n=0}^{\infty}x^{2n}, \qquad B(x)=-\sum_{n=0}^{\infty}x^{2n} +\sum_{n=0}^{\infty}\left(\frac{x}{3}\right)^n.

The coefficients of \(A\) are \(1\) in even degrees and \(0\) in odd degrees. Their root limit superior is \(1\), so \(A\) has radius \(1\). For \(B\), the coefficient of \(x^k\) is \(-1+3^{-k}\) when \(k\) is even and \(3^{-k}\) when \(k\) is odd. Along even degrees, the coefficient tends to \(-1\), and the \(k\)th root of its absolute value tends to \(1\). Along odd degrees, the coefficient root tends to \(1/3\). The limit superior across all degrees is therefore \(1\), so \(B\) also has radius \(1\).

In the coefficientwise sum, the terms \(\sum x^{2n}\) and \(-\sum x^{2n}\) cancel in every degree. What remains is

A(x)+B(x)=\sum_{n=0}^{\infty}\left(\frac{x}{3}\right)^n,

whose radius is \(3\). To check the coefficient alignment at \(x=1/2\), the individual series converge because \(|x|<1\):

A\left(\frac12\right)=\frac{1}{1-1/4}=\frac43, \qquad B\left(\frac12\right)=-\frac43+\frac{1}{1-1/6} =-\frac43+\frac65=-\frac{2}{15}.

Their sum is \(4/3-2/15=20/15-2/15=18/15=6/5\). The remaining series gives \(1/(1-1/6)=6/5\) as well, confirming the coefficientwise result.

A Common Mistake

A frequent error is to assume that the radius of the sum must equal the smaller original radius. For the cancellation example above, that incorrect calculation would be

R_{\text{sum}}=\min(1,1)=1.

This treats the theorem's guarantee as an equality. But the actual sum has coefficients \(2^{-n}\), whose root limit is \(1/2\), so its radius is \(2\). The correct conclusion from the theorem alone is \(R_{\text{sum}}\geq1\); to find the exact radius, calculate the combined coefficients and analyze them.

A related mistake is to add the values of two series at a point where one or both do not converge. Coefficientwise cancellation can produce a convergent sum there, but it does not give the divergent original series separate values. Keep the coefficient calculation distinct from adding function values outside the common convergence region.

What the Result Does—and Does Not—Say

The lower bound applies to power series with the same center, added degree by degree. It requires no cancellation assumption and follows just from absolute convergence and the triangle inequality. It also gives no guarantee that the sum has a boundary behavior matching either original series. The endpoint questions remain separate from the radius calculation.

Claim	What is justified
The sum's radius is at least the smaller radius.	Always, for two power series with a common center.
The sum's radius equals the smaller radius.	Not always; determine the combined coefficients to know.
The sum's radius can be larger.	Yes, if coefficient cancellation improves their rate of decrease.
Boundary convergence follows from the radius.	No; test each boundary point separately.

The essential procedure is simple: align the coefficients at the common center, add them, and then distinguish the guaranteed lower bound from the actual radius. The theorem protects convergence inside the smaller radius; cancellation explains why the sum can sometimes continue farther.

Check Your Understanding

Use the theorem and coefficient calculations where appropriate.

If two power series with the same center have radii \(4\) and \(7\), what convergence is guaranteed for their coefficientwise sum?
Why does the triangle inequality prove absolute convergence inside both original radii?
For coefficients \(a_n=1\) and \(b_n=-1+5^{-n}\), find the radii of the two original series and of their coefficientwise sum.
Can the theorem alone determine whether the sum converges at a point exactly one smaller radius from the center? Explain.
Why is it incorrect to add the values of two divergent series at a point merely because their coefficientwise sum converges there?