Sparse Coefficients and Radius Calculation

Why Sparse Coefficients Need Care

A power series may have many zero coefficients. For instance, a series involving only even powers has zero coefficients at every odd index. This pattern can make the ordinary ratio of consecutive coefficients unhelpful: ratios involving a zero may be undefined, and ratios across gaps may alternate or fail to settle to a limit.

The radius of convergence is still determined by the coefficients. The key is to use the limit superior of their $n$th roots, with $n$ equal to the actual power of $x$. If the nonzero terms occur at powers $n_1,n_2,\ldots$, the relevant roots are $\lvert a_{n_j}\rvert^{1/n_j}$, not roots indexed by the position $j$ in that list.

Definition: For a power series $\sum_{n=0}^{\infty}a_nx^n$, let $S=\{n\in\mathbb N:a_n\ne 0\}$ be its positive nonzero indices. If $S$ is infinite, define $$ L=\limsup_{\substack{n\to\infty\\ n\in S}} |a_n|^{1/n}. $$ The radius of convergence is $R=1/L$, with the conventions $1/0=\infty$ and $1/\infty=0$. If there are only finitely many nonzero coefficients, the series is a polynomial and its radius is $\infty$.

Including or omitting a finite number of indices, such as the constant coefficient $a_0$, does not affect the limit superior. Zero coefficients between the nonzero indices do not contribute to the sparse-index calculation.

Why the Limit Superior Gives the Radius

The Cauchy–Hadamard formula expresses the radius using all the coefficients: $R=1/\limsup_{n\to\infty}|a_n|^{1/n}$. For a sparse series, its limit superior is exactly the limit superior taken along its nonzero indices. The following argument also explains how that number determines convergence away from the boundary.

Take the root of a term.
For $x\ne 0$, the $n$th root of the absolute value of the $n$th term is $|a_nx^n|^{1/n}=|a_n|^{1/n}|x|$. At a zero coefficient this root is zero.

Take the limit superior.
The limit superior of these roots is $L|x|$, where $L$ is the limit superior along the nonzero indices. Multiplication by the fixed positive number $|x|$ multiplies the limit superior by $|x|$.

Apply the root test.
If $L|x|<1$, the root test gives absolute convergence. If $L|x|>1$, the terms do not tend to zero, so the series diverges.

Thus all points with $|x|<1/L$ are inside the radius, and all points with $|x|>1/L$ are outside it. When $L=0$, every finite $x$ gives $L|x|=0<1$, so the radius is infinite. The equality case $L|x|=1$ is not decided by this calculation; the radius alone does not determine convergence at boundary points.

Worked Examples

Worked Example: Nonzero Coefficients at Even Powers

Find the radius of the series $\sum_{k=1}^{\infty}9^k x^{2k}$. Its nonzero coefficients are $a_{2k}=9^k$, and all odd-index coefficients are zero. The actual exponent of the $k$th nonzero term is $2k$, so calculate

|a_{2k}|^{1/(2k)} =(9^k)^{1/(2k)} =9^{1/2} =3.

This value is $3$ for every $k$, so the limit superior is $L=3$. The radius is therefore

R=\frac{1}{3}.

Check the result using the root test. At $x=1/6$, the root along the nonzero terms is $3|1/6|=1/2<1$, so the series converges absolutely. At $x=1/2$, it is $3|1/2|=3/2>1$, so the series diverges. These checks confirm that the calculated radius separates points inside from points outside it.

Worked Example: Coefficients at Square Powers

Consider $\sum_{k=1}^{\infty}3^k x^{k^2}$. The nonzero coefficients occur at the square indices $n=k^2$, where $a_{k^2}=3^k$. The root must use the exponent $k^2$:

|a_{k^2}|^{1/k^2} =(3^k)^{1/k^2} =3^{1/k}.

As $k\to\infty$, $3^{1/k}\to 1$. Therefore $L=1$ and

R=\frac{1}{1}=1.

For a check, at $x=1/2$ the root of the absolute value of the $k$th nonzero term is $(3^k(1/2)^{k^2})^{1/k^2}=3^{1/k}/2$, which tends to $1/2<1$; the series converges absolutely there. At $x=2$, the corresponding root is $2\cdot3^{1/k}$, which tends to $2>1$; the series diverges. Both checks agree with radius $1$.

Worked Example: Nonzero Coefficients at Powers of Two

Find the radius of $\sum_{j=1}^{\infty}5^{2^j}x^{2^j}$. Here the nonzero indices are $n=2^j$, and the coefficient at that index is $a_{2^j}=5^{2^j}$. Thus

|a_{2^j}|^{1/2^j} =(5^{2^j})^{1/2^j} =5.

The limit superior is $L=5$, so $R=1/5$. To check, at $x=1/10$ the root of each nonzero term is $5(1/10)=1/2<1$, giving absolute convergence. At $x=3/10$ it is $5(3/10)=3/2>1$, giving divergence. The radius $1/5$ is consistent with both tests.

The Index in the Root Matters

When listing nonzero terms, it is natural to number them $1,2,3,\ldots$. That numbering describes their positions in the list, not their powers in the series. The Cauchy–Hadamard formula uses the power $n$. For nonzero indices $n_j$, it therefore uses $|a_{n_j}|^{1/n_j}$, not $|a_{n_j}|^{1/j}$.

The square-power example shows why this distinction matters. There, the coefficients listed in order are $3,3^2,3^3,\ldots$, so taking a $k$th root gives $3$. But the coefficient $3^k$ belongs to $x^{k^2}$, not $x^k$. Its correct root is $(3^k)^{1/k^2}=3^{1/k}$, which tends to $1$. Using the list position as the root index would produce the wrong radius.

Common Pitfall: Expecting the Ratio to Behave Regularly

For a series with coefficients only at even powers, the coefficient sequence contains repeated zeros between nonzero entries. For example, in $\sum_{k=1}^{\infty}9^kx^{2k}$, $a_2=9$, $a_3=0$, and $a_4=81$. The ratio $a_3/a_2=0$, while $a_4/a_3$ is undefined. The ordinary consecutive-coefficient ratio therefore does not produce a useful limit.

A tempting but incorrect response is to use only the nonzero coefficients $9,81,729,\ldots$ and calculate consecutive ratios $81/9=9$, then infer a radius $1/9$. That ratio treats the list as though its entries belonged to consecutive powers $x,x^2,x^3,\ldots$. They actually belong to $x^2,x^4,x^6,\ldots$. The correct roots are $(9^k)^{1/(2k)}=3$, giving $R=1/3$. The sparse-index limit superior respects the actual powers and avoids this mismatch.

Why This Calculation Is Useful

Sparse power series occur whenever a function or construction uses only selected powers—for example, only even powers or only powers with square indices. The gaps do not remove the need to account for the degree of each term. The root formula handles these gaps directly, even when consecutive coefficient ratios fail to exist or have no limit.

For a practical calculation, first identify the actual nonzero indices $n_j$. Then write the coefficient at each such index as $a_{n_j}$, form $|a_{n_j}|^{1/n_j}$, and find its limit superior. Finally, take the reciprocal, using the stated conventions if the limit superior is zero or infinite. This gives the radius, while separate tests are needed to decide convergence at its boundary.

Check Your Understanding

For each question, pay attention to the exponent of $x$, not just to the order in which the nonzero terms are listed.

A power series has $a_{2k}=16^k$ and all odd-index coefficients zero. What limit superior determines its radius, and what is the radius?
For a series with $a_{k^2}=7^k$, what sequence of roots should be used? What does it tend to?
Explain why the ordinary ratio of consecutive coefficients may fail to determine the radius when coefficients are sparse.
A series has $a_{2^j}=4^{2^j}$ and all other positive-index coefficients zero. Find its radius using the limit superior.
If the limit superior of the roots along the nonzero indices is zero, what is the radius, and what does the root test say for finite $x$?