§2 Remainders and Power Series

Knowing that a series converges does not tell us its exact sum or how many terms are needed for a useful approximation. Occasionally an exact identity is known, while a similar-looking series may have no immediately useful formula:

$$\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}, \qquad \sum_{n=1}^{\infty}\frac{1}{n^3}=?$$

Remainder estimates put a guaranteed bound on the part of a series we did not add, so a finite partial sum can still give a controlled approximation.

The same convergence tools also let us study power series, whose terms depend on a variable $x$. A power series may converge for some values of $x$ and diverge for others.

Partial sums and remainders

Suppose the convergent series

$$S=\sum_{n=0}^{\infty}a_n$$

has $N$th partial sum

$$S_N=\sum_{n=0}^{N}a_n.$$

The remainder, also called the tail or error, is

$$\boxed{R_N=S-S_N=\sum_{n=N+1}^{\infty}a_n}.$$

The first omitted term is $a_{N+1}$ because $S_N$ already includes every term from $a_0$ through $a_N$.

A useful remainder estimate answers a practical question:

How far can the approximation $S_N$ be from the true sum $S$?

Integral-test remainder estimate

Suppose $f$ is continuous, positive, and decreasing, and

$$a_n=f(n).$$

If the corresponding improper integral—an integral whose interval extends to infinity—converges, then the remainder satisfies

$$\boxed{ \int_{N+1}^{\infty}f(x)\,dx \le R_N \le \int_N^{\infty}f(x)\,dx. }$$

The left integral is a lower bound and the right integral is an upper bound.

Why do the bounds begin at different places?

Because $f$ is decreasing, its value across a unit interval can be compared with a rectangle whose height is one of the sequence terms.

For the lower bound, pair $a_n=f(n)$ with the interval immediately to its right:

$$\int_n^{n+1}f(x)\,dx\le f(n)=a_n.$$

The remainder begins with $a_{N+1}$, so the first interval is $[N+1,N+2]$. Adding all those inequalities gives

$$\int_{N+1}^{\infty}f(x)\,dx\le R_N.$$

For the upper bound, pair $a_n=f(n)$ with the interval immediately to its left:

$$f(n)=a_n\le\int_{n-1}^{n}f(x)\,dx.$$

The interval paired with $a_{N+1}$ begins at $N$, so

$$R_N\le\int_N^{\infty}f(x)\,dx.$$

The shift by $1$ is not arbitrary. It comes from aligning the first omitted term with a whole unit interval. Starting at $N+\tfrac12$ would require extra assumptions; positivity and monotonic decrease alone do not guarantee that comparison.

Using an upper bound to choose $N$

If we want an error smaller than a tolerance $\varepsilon>0$, it is enough to choose $N$ so that

$$\int_N^{\infty}f(x)\,dx<\varepsilon.$$

Then

$$|S-S_N|=R_N<\varepsilon.$$

For a positive series, $R_N$ is positive, so absolute values are optional. They become essential for alternating series.

Alternating-series remainder estimate

Consider an alternating series

$$S=\sum_{n=0}^{\infty}(-1)^n b_n,$$

where

$$b_n\ge0, \qquad b_{n+1}\le b_n, \qquad \lim_{n\to\infty}b_n=0.$$

These are the hypotheses of the Alternating Series Test, abbreviated AST. They guarantee convergence. They also give the stronger error estimate

$$\boxed{|R_N|=|S-S_N|\le b_{N+1}}.$$

In words: the approximation error is no larger than the magnitude of the first omitted term.

Example: enough accuracy to round correctly

Consider

$$S=\sum_{n=0}^{\infty} \frac{(-1)^n}{4^n(n+1)}.$$

Here

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

The terms are positive, decrease, and approach zero, so the AST and its remainder estimate apply.

Using terms through $n=3$ gives

$$\begin{aligned} S_3 &=1-\frac18+\frac1{48}-\frac1{256}\\ &=0.891927083\ldots. \end{aligned}$$

The first omitted magnitude is

$$b_4=\frac{1}{4^4(5)}=\frac1{1280}=0.00078125.$$

Therefore,

$$|S-S_3|\le0.00078125.$$

Even after adding or subtracting the maximum possible error, the result still rounds to

$$\boxed{S\approx0.89}$$

to two decimal places.

Stopping at $N=2$ would not be enough. The allowed error interval would contain values that round to $0.89$ and values that round to $0.90$. A decimal approximation is useful only when the error bound is small enough to settle the requested rounding.

Power series

A power series centered at $a$ is an infinite series of the form

$$\boxed{ \sum_{n=0}^{\infty}c_n(x-a)^n =c_0+c_1(x-a)+c_2(x-a)^2+\cdots. }$$

Here:

• $x$ is the variable;
• $a$ is a fixed real number called the center;
• $c_n$ is the coefficient of the $n$th power.

When $a=0$, this becomes

$$\sum_{n=0}^{\infty}c_nx^n.$$

A power series resembles a polynomial, but a polynomial has only finitely many nonzero terms. A power series may have infinitely many.

Radius and interval of convergence

For a power series

$$\sum_{n=0}^{\infty}c_n(x-a)^n,$$

exactly one of the following occurs:

1. It converges only at $x=a$.
2. It converges for every real $x$.
3. There is a number $R>0$ such that it converges when $|x-a|<R$ and diverges when $|x-a|>R$.

The number $R$ is the radius of convergence. By convention:

• $R=0$ in the first case;
• $R=\infty$ in the second case.

If $0<R<\infty$, then

$$|x-a|<R \quad\Longleftrightarrow\quad a-R<x<a+R.$$

The points $x=a-R$ and $x=a+R$ are the endpoints. The radius theorem says nothing about them, so they must be checked separately.

The interval of convergence is the complete set of real $x$-values where the series converges, including whichever endpoints pass their individual tests.

Example: the endpoints behave differently

Find the interval of convergence of

$$\sum_{n=1}^{\infty}\frac{(x-3)^n}{n}.$$

Let

$$a_n=\frac{(x-3)^n}{n}.$$

First handle $x=3$: every term is zero, so the series converges. For $x\ne3$, the ratio test gives

$$\begin{aligned} \left|\frac{a_{n+1}}{a_n}\right| &=\left|(x-3)\frac{n}{n+1}\right|\\ &\longrightarrow |x-3|. \end{aligned}$$

The series therefore converges absolutely, meaning that the series of absolute values also converges, when

$$|x-3|<1,$$

which means

$$2<x<4.$$

Now check the endpoints.

At $x=2$,

$$\sum_{n=1}^{\infty}\frac{(-1)^n}{n}$$

is an alternating harmonic series, so it converges conditionally: it converges, but the corresponding series of absolute values does not.

At $x=4$,

$$\sum_{n=1}^{\infty}\frac1n$$

is the harmonic series, so it diverges.

Thus the center is $3$, the radius is $1$, and the interval of convergence is

$$\boxed{[2,4)}.$$

In set-builder notation,

$$[2,4)=\{x\in\mathbb R:2\le x<4\}.$$

A power series defines a function

Inside its interval of convergence, a power series defines a function:

$$f(x)=\sum_{n=0}^{\infty}c_n(x-a)^n.$$

Sometimes the same function has a simpler formula. The geometric-series identity says

$$\frac{1}{1-u}=\sum_{n=0}^{\infty}u^n, \qquad |u|<1.$$

The restriction belongs to the series representation. The formula $1/(1-u)$ may be defined at points where the series diverges.

Example: build a representation from a geometric series

Find a power-series representation centered at $0$ for

$$f(x)=\frac{3}{2+x^2}.$$

Rewrite the denominator to match $1/(1-u)$:

$$\begin{aligned} \frac{3}{2+x^2} &=\frac32\frac{1}{1+x^2/2}\\ &=\frac32\frac{1}{1-(-x^2/2)}. \end{aligned}$$

Use

$$u=-\frac{x^2}{2}.$$

Then

$$\begin{aligned} \frac{3}{2+x^2} &=\frac32\sum_{n=0}^{\infty} \left(-\frac{x^2}{2}\right)^n\\ &=\boxed{ \sum_{n=0}^{\infty} \frac{3(-1)^n}{2^{n+1}}x^{2n} }. \end{aligned}$$

This converges when

$$\left|-\frac{x^2}{2}\right|<1,$$

so

$$|x|<\sqrt2.$$

At $x=\pm\sqrt2$, the resulting terms do not approach zero. Therefore the interval of convergence is

$$\boxed{(-\sqrt2,\sqrt2)}.$$

Only even powers appear. For example, the $x^6$ term occurs when

$$2n=6,$$

so it comes from $n=3$, not $n=6$.

Differentiating and integrating power series

Suppose

$$f(x)=\sum_{n=0}^{\infty}c_n(x-a)^n$$

has radius of convergence $R>0$. Inside the open interval

$$(a-R,a+R),$$

we may differentiate and integrate term by term.

Differentiation

$$\begin{aligned} f'(x) &=c_1+2c_2(x-a)+3c_3(x-a)^2+\cdots\\ &=\sum_{n=1}^{\infty}n c_n(x-a)^{n-1}\\ &=\sum_{n=0}^{\infty}(n+1)c_{n+1}(x-a)^n. \end{aligned}$$

The final line is the same series after reindexing, meaning that the summation label is changed so the exponent is again $n$.

Integration

$$\begin{aligned} \int f(x)\,dx &=C+c_0(x-a)+\frac{c_1}{2}(x-a)^2 +\frac{c_2}{3}(x-a)^3+\cdots\\ &=C+\sum_{n=0}^{\infty} \frac{c_n}{n+1}(x-a)^{n+1}. \end{aligned}$$

The differentiated and integrated series have the same radius of convergence $R$. Their behavior at the endpoints may change, so endpoint convergence must still be checked separately.

Example: differentiating the geometric series

Starting from

$$\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n, \qquad |x|<1,$$

differentiate both sides:

$$\boxed{ \frac{1}{(1-x)^2} =\sum_{n=1}^{\infty}n x^{n-1} =\sum_{n=0}^{\infty}(n+1)x^n }, \qquad |x|<1.$$

Example: the exponential series

Define

$$f(x)=\sum_{n=0}^{\infty}\frac{x^n}{n!}.$$

Term-by-term differentiation gives

$$\begin{aligned} f'(x) &=\sum_{n=1}^{\infty}n\frac{x^{n-1}}{n!}\\ &=\sum_{n=1}^{\infty}\frac{x^{n-1}}{(n-1)!}\\ &=\sum_{n=0}^{\infty}\frac{x^n}{n!}\\ &=f(x). \end{aligned}$$

Also $f(0)=1$. The function with

$$f'=f, \qquad f(0)=1$$

is $e^x$. Therefore,

$$\boxed{e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}}$$

for every real $x$.

A practical power-series workflow

To find an interval of convergence:

1. Identify the center $a$.
2. Apply the ratio or root test to find the condition $|x-a|<R$.
3. Convert that inequality into an open interval.
4. Test the two endpoints separately.
5. Use brackets only for endpoints that actually converge.

To approximate a convergent numerical series:

1. Decide which remainder estimate applies.
2. Express the bound in terms of $N$.
3. Choose $N$ so the bound is small enough for the requested accuracy.
4. Compute $S_N$.
5. Report the approximation together with its guaranteed error.