Substituting the Two Endpoints Into a Power Series

Why Endpoint Substitution Needs Care

A power series centered at $c$ is written in powers of the distance $x-c$. Once its radius $R$ is known, the two boundary points are $c+R$ and $c-R$. Substitution at these points has a simple algebraic form, but the signs differ: the right endpoint contributes $R^n$, whereas the left endpoint contributes $(-1)^nR^n$.

That distinction matters because a coefficient may already contain a factor of $(-1)^n$, or another sign pattern. At the left endpoint, the sign from the substitution must be multiplied by the sign in the coefficient. Writing the two resulting numerical series correctly is the goal here. Whether either series converges is a separate question.

Definition: A power series centered at $c$ has the form $\sum_{n=0}^{\infty}a_n(x-c)^n$, where the $a_n$ are its coefficients. If its radius of convergence is a finite positive number $R$, its boundary points are $c+R$ and $c-R$.

The radius describes how far the boundary points are from the center. It does not say that the series converges at either boundary point. To examine a boundary point, substitute that value of $x$ into each term and simplify.

The Two Substitution Formulas

For a term $a_n(x-c)^n$, substitution at the right endpoint gives

x=c+R \quad\Longrightarrow\quad x-c=(c+R)-c=R, \qquad a_n(x-c)^n=a_nR^n.

At the left endpoint, the displacement from the center is negative:

x=c-R \quad\Longrightarrow\quad x-c=(c-R)-c=-R.

For every nonnegative integer $n$, $(-R)^n=(-1)^nR^n$. Therefore, at the left endpoint, the term becomes $a_n(-1)^nR^n$. This factor $(-1)^n$ alternates between $1$ for even $n$ and $-1$ for odd $n$.

Substitution rule: For $\sum_{n=0}^{\infty}a_n(x-c)^n$ with finite positive radius $R$, the formal boundary series are $$ \begin{aligned} x=c+R &: \quad \sum_{n=0}^{\infty}a_nR^n,\\ x=c-R &: \quad \sum_{n=0}^{\infty}a_n(-1)^nR^n. \end{aligned} $$ These formulas identify the series to be examined; by themselves, they do not assert that either series converges.

The formulas follow directly by evaluating the terms. To be precise about the infinite series, consider its partial sum through index $N$, which is a finite sum and can always be evaluated by substitution. At $x=c+R$, it becomes $\sum_{n=0}^{N}a_nR^n$; at $x=c-R$, it becomes $\sum_{n=0}^{N}a_n(-1)^nR^n$. Taking the limit of these partial sums, if that limit exists, is exactly what it means for the original series to converge at that endpoint.

Deriving the Formulas from Partial Sums

The partial-sum argument both justifies the formulas and explains their purpose: endpoint substitution turns a power series into a numerical series, with no variable remaining.

Start with a finite partial sum.
Write $S_N(x)=\sum_{n=0}^{N}a_n(x-c)^n$. Because the sum is finite, substitution distributes across its terms.

Substitute the right endpoint.
Putting $x=c+R$ gives $x-c=R$, so $S_N(c+R)=\sum_{n=0}^{N}a_nR^n$.

Substitute the left endpoint.
Putting $x=c-R$ gives $x-c=-R$, so $S_N(c-R)=\sum_{n=0}^{N}a_n(-R)^n=\sum_{n=0}^{N}a_n(-1)^nR^n$.

Interpret the resulting series.
At either endpoint, the power series converges precisely when the corresponding sequence of partial sums has a finite limit. The substitution formulas do not determine whether that limit exists.

This derivation also shows why it is best to substitute into the whole term rather than try to guess the endpoint pattern. The power $(x-c)^n$ carries the endpoint sign, while $a_n$ may carry signs of its own.

Worked Examples

Worked Example: A Series Centered Away from Zero

Consider $\sum_{n=1}^{\infty}\frac{(x-2)^n}{n3^n}$, whose center is $c=2$ and whose radius is $R=3$. Find the series obtained at each endpoint.

The boundary points are $c+R=2+3=5$ and $c-R=2-3=-1$. The general term is $\frac{(x-2)^n}{n3^n}$.

At the right endpoint, $x=5$, so $x-2=3$. For each index $n$,

\frac{(5-2)^n}{n3^n} = \frac{3^n}{n3^n} = \frac{1}{n}.

Thus the right-endpoint series is $\sum_{n=1}^{\infty}\frac{1}{n}$.

At the left endpoint, $x=-1$, so $x-2=-3$. Then

\frac{(-1-2)^n}{n3^n} = \frac{(-3)^n}{n3^n} = \frac{(-1)^n3^n}{n3^n} = \frac{(-1)^n}{n}.

Thus the left-endpoint series is $\sum_{n=1}^{\infty}\frac{(-1)^n}{n}$.

Check: The endpoint displacements are $5-2=3$ and $-1-2=-3$. Their $n$th powers divided by $3^n$ give $1$ and $(-1)^n$, respectively, as in the two displayed series.

Worked Example: Combining Two Alternating Factors

Consider $\sum_{n=0}^{\infty}\frac{(-1)^n(x-1)^n}{2^n(n+1)}$, with center $c=1$ and radius $R=2$. Find both endpoint series, paying attention to the sign already present in the coefficient.

The endpoints are $1+2=3$ and $1-2=-1$. At $x=3$, the term simplifies as follows:

\frac{(-1)^n(3-1)^n}{2^n(n+1)} = \frac{(-1)^n2^n}{2^n(n+1)} = \frac{(-1)^n}{n+1}.

The right-endpoint series is therefore $\sum_{n=0}^{\infty}\frac{(-1)^n}{n+1}$.

At $x=-1$, the displacement is $-1-1=-2$. The coefficient sign and the displacement sign must both be included:

\begin{aligned} \frac{(-1)^n(-1-1)^n}{2^n(n+1)} &= \frac{(-1)^n(-2)^n}{2^n(n+1)}\\ &= \frac{(-1)^n(-1)^n2^n}{2^n(n+1)}\\ &= \frac{1}{n+1}. \end{aligned}

Here $(-1)^n(-1)^n=(-1)^{2n}=1$, since $2n$ is even. The left-endpoint series is $\sum_{n=0}^{\infty}\frac{1}{n+1}$.

Check: Substituting $x=-1$ gives two factors of $(-1)^n$, one from the coefficient and one from $(-2)^n$. Their product is $1$ for every $n$, confirming the nonalternating terms in the left-endpoint series.

Worked Example: The Radius Factor Cancels a Coefficient Factor

Consider $\sum_{n=0}^{\infty}\frac{3^n}{n^2+1}(x-4)^n$. Its center is $c=4$. The ratio of consecutive coefficient magnitudes is

\frac{\frac{3^{n+1}}{(n+1)^2+1}}{\frac{3^n}{n^2+1}} = 3\frac{n^2+1}{(n+1)^2+1} \longrightarrow 3.

Thus its radius is $R=\frac{1}{3}$. The endpoints are $4+\frac{1}{3}=\frac{13}{3}$ and $4-\frac{1}{3}=\frac{11}{3}$.

At the right endpoint, $\frac{13}{3}-4=\frac{1}{3}$. The $n$th term becomes

\frac{3^n}{n^2+1}\left(\frac{1}{3}\right)^n = \frac{3^n}{n^2+1}\frac{1}{3^n} = \frac{1}{n^2+1}.

So the right-endpoint series is $\sum_{n=0}^{\infty}\frac{1}{n^2+1}$.

At the left endpoint, $\frac{11}{3}-4=-\frac{1}{3}$. This gives

\frac{3^n}{n^2+1}\left(-\frac{1}{3}\right)^n = \frac{3^n}{n^2+1}\frac{(-1)^n}{3^n} = \frac{(-1)^n}{n^2+1}.

So the left-endpoint series is $\sum_{n=0}^{\infty}\frac{(-1)^n}{n^2+1}$.

Check: In both cases, the factor $3^n$ in the coefficient cancels the magnitude $3^n$ in the denominator from the endpoint displacement. The negative displacement at the left endpoint leaves exactly the additional factor $(-1)^n$.

A Common Sign Error

A frequent mistake is to treat the left endpoint as though its displacement were $R$, rather than $-R$. For the first example, this mistake would calculate

\frac{(-1-2)^n}{n3^n} \ \color{red}{=}\ \frac{3^n}{n3^n} = \frac{1}{n}.

The equality marked in red is wrong: $-1-2=-3$, not $3$. The corrected calculation is

\frac{(-1-2)^n}{n3^n} = \frac{(-3)^n}{n3^n} = \frac{(-1)^n}{n}.

The reliable correction is to write the displacement first, then raise it to the $n$th power. In general, record $x-c=-R$ at the left endpoint and use $(-R)^n=(-1)^nR^n$. Do not decide the final sign until you have also included any sign already in $a_n$.

What the Substitution Does—and Does Not—Tell You

After substitution, each endpoint has a numerical series that can be studied using appropriate tests. But the algebraic conversion and the convergence decision are different tasks. For example, obtaining $\sum_{n=1}^{\infty}\frac{(-1)^n}{n}$ identifies the left-endpoint series; it does not, merely by being written down, prove that the series converges.

Keeping these tasks separate prevents two common errors: claiming an endpoint is included just because it is exactly one radius from the center, and assuming that both endpoint series must behave the same way. The signs, coefficients, and resulting terms can differ. First derive each series accurately; then analyze the two endpoints on their own.

Takeaway: At $c+R$, replace $(x-c)^n$ with $R^n$. At $c-R$, replace it with $(-1)^nR^n$. Combine that endpoint factor with the full coefficient $a_n$, and only then proceed to a convergence analysis.

Check Your Understanding

For each question, focus first on the algebra of substitution rather than on deciding convergence.

For $\sum_{n=0}^{\infty}a_n(x-c)^n$ with radius $R$, what term replaces $a_n(x-c)^n$ at $x=c+R$? What term replaces it at $x=c-R$?
A series is centered at $5$ and has radius $2$. What are its two boundary points, and what are the corresponding displacements $x-c$?
At the left endpoint, a coefficient is $a_n=(-1)^nb_n$. Write the endpoint term in terms of $b_n$ and $R^n$, simplifying the two sign factors.
For $\sum_{n=0}^{\infty}\frac{(x-3)^n}{4^n(n+1)}$, write the numerical series produced at each boundary point.
Does correctly writing an endpoint series, on its own, establish that the original power series converges at that endpoint? Explain briefly.