Walkthrough

Remainders

An infinite series represents a sum that never ends:

Because we cannot add terms forever, we must stop at some finite cutoff index $N$. This splits the series into two parts:

1. The partial sum $S_N = a_0 + a_1 + \cdots + a_N$, which is our calculated approximation.
2. The remainder $R_N = a_{N+1} + a_{N+2} + \cdots$, which is the leftover tail we omitted.

A common point of confusion is what the index $N$ represents. The subscript $N$ marks the last included index, not the total count of terms. If the series starts at $n = 0$, the partial sum $S_3$ contains four terms ($a_0, a_1, a_2, a_3$), and the remainder $R_3$ starts at index $N+1 = 4$.

To know whether an approximation is reliable, we need to bound the size of this discarded remainder.

---

Estimating the remainder

How we estimate $R_N$ depends on the behavior of the terms.

Integral estimate

For a convergent series $S=\sum_{n=0}^{\infty}a_n$ and an integer $N\ge0$, suppose $a_n=f(n)$ for $n\ge N$ and $f$ is continuous, positive, and nonincreasing on $[N,\infty)$. Then we can trap the omitted tail between areas under $y=f(x)$:

• Starting the integral at $N+1$ places the curve beneath the rectangles, giving a lower bound.
• Starting the integral at $N$ shifts the reference frame so the curve stays above the rectangles, giving an upper bound.

If you are asked to guarantee that the error remains below a chosen tolerance $\varepsilon$, focus on the upper bound: find an index $N$ such that $\int_N^{\infty} f(x)\,dx < \varepsilon$. Because $R_N$ is smaller than this integral, your error is guaranteed to be less than $\varepsilon$.

Alternating-series estimate

For an alternating series whose magnitudes $b_n\ge0$ are nonincreasing and approach $0$, the Alternating Series Estimation Theorem bounds the truncation error by the first omitted magnitude:

To estimate the true sum to a specific precision, calculate $S_N$ and check whether the window $[S_N - b_{N+1}, S_N + b_{N+1}]$ rounds unambiguously to your target decimal place.

---

Power series

A power series extends a polynomial to infinitely many terms and is centered at a constant $a$:

At the center $x = a$, every term with $n \ge 1$ becomes zero, leaving only $c_0$. Therefore, every power series converges at its center.

For other values of $x$, convergence depends on how far $x$ is from the center. This behavior is captured by the radius of convergence, $R$:

• The series converges absolutely when $|x - a| < R$.
• The series diverges when $|x - a| > R$.

Interval of convergence

To find the complete set of $x$-values where the series converges:

1. Use the Ratio Test—or the Root Test when it fits better—to determine the interior condition $|x-a|<R$.
2. Rewrite that condition as the open interval $(a-R,a+R)$.
3. When $R$ is finite, test each endpoint individually by substituting $x=a-R$ and $x=a+R$ into the original series. The test used to find the radius is generally inconclusive at the boundary, so a different numerical-series test decides whether each endpoint is included.

---

Operations on power series

Inside the open interval of convergence, power series can be treated like polynomials:

• Algebraic substitutions: A rational function can often be reshaped into $\frac{A}{1-u}$ to use the geometric series formula $\frac{1}{1-u} = \sum_{n=0}^{\infty} u^n$ for $|u| < 1$.
• Calculus term by term: You can differentiate and integrate a power series term by term. The radius of convergence $R$ remains unchanged, though endpoint behavior must be re-evaluated.

For example, differentiating $\sum_{n=0}^{\infty} \frac{x^n}{n!}$ term by term returns the same series, and its value at $x=0$ is $1$. These are the defining properties behind the familiar Maclaurin series for $e^x$.

Partial sums and remainders

A series contains infinitely many terms, but in practice we can only add finitely many. That limitation raises two practical questions:

1. If we truncate a series, how large can the omitted part be?
2. If the terms depend on an input variable $x$, for which values of $x$ does the series converge?

While some series evaluate to a known closed form, such as

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

other closely related series have no simple exact formula:

$$\sum_{n=1}^{\infty}\frac{1}{n^3}=?$$

Error estimates allow us to quantify how closely a finite computation approximates the true infinite sum.

Example: $N=3$

Consider an infinite series written term by term from an initial index $n=0$:

$$S=a_0+a_1+a_2+a_3+a_4+a_5+\cdots.$$

If we stop after the term with index $3$, we obtain the partial sum

$$S_3=a_0+a_1+a_2+a_3.$$

The difference between the total sum $S$ and this finite approximation is the remainder $R_3$:

$$R_3=S-S_3=a_4+a_5+a_6+\cdots.$$

Because $a_3$ is already included inside $S_3$, the first missing term in the remainder is $a_4$.

Because this series begins at $n=0$, the partial sum $S_3$ contains four terms. The subscript $N$ denotes the last included index, not the total number of terms added.

More generally, stopping at index $N$ produces

$$S_N=a_0+a_1+\cdots+a_N$$

and leaves the remainder

$$\boxed{ R_N=S-S_N=a_{N+1}+a_{N+2}+\cdots =\sum_{n=N+1}^{\infty}a_n }.$$

The remainder $R_N$ is the signed leftover between the true sum and the approximation. The error size is its absolute value, $|R_N|$. When every term $a_n$ is positive, $R_N>0$, so the remainder and the error size coincide. For alternating series, $R_N$ can be positive or negative, requiring us to bound $|R_N|$.

Integral remainder bounds

Integral remainder estimate

When the terms of a series are generated by a continuous function,

$$a_n=f(n),$$

we can bound the remainder geometrically using areas under the curve. For this comparison to hold, $f$ must satisfy three conditions:

• continuous, ensuring the graph has no jumps or breaks;
• positive, ensuring the area lies entirely above the $x$-axis;
• decreasing, ensuring successive terms shrink in size.

The following diagram illustrates these comparisons for $N=3$, starting from $n=0$:

Lower bound for $R_3$

With $N=3$, the first four rectangles are already included in $S_3$. The remainder starts at $N+1=4$.

$S_3=a_0+a_1+a_2+a_3$: already included

$R_3=a_4+a_5+a_6+\cdots$: first omitted term onward

$$\displaystyle \int_4^\infty f(x)\,dx\le R_3$$

From $x=4$ onward, the curve lies below the tail rectangles. Its area is therefore a lower estimate of the tail.

Upper bound for $R_3$

The gray rectangles stay in place for context. The first tail rectangle is still $a_4$, but placing it over $[3,4]$ makes the curve sit above it.

$$\displaystyle R_3\le \int_3^\infty f(x)\,dx$$

Starting the area at $x=3$ puts the curve above the same tail rectangles, so this area is an upper estimate.

To see where the bounds come from, compare an individual term $a_n=f(n)$—represented by a rectangle of width $1$ and height $f(n)$—with the area under $f(x)$ over unit intervals:

• Comparing to the unit interval to the right gives

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

• Comparing to the unit interval to the left gives

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

Returning to the $N=3$ diagram, the remainder tail starts at $n=4$:

$$R_3=a_4+a_5+a_6+\cdots.$$

Summing the single-rectangle inequalities from $n=N+1$ to $\infty$ yields the integral remainder estimate:

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

The difference in limits reflects the boundary between the included and omitted terms:

$$S_N\text{ ends at }a_N, \qquad R_N\text{ begins at }a_{N+1}.$$

• Integrating from $x=4$ places the curve below the tail rectangles, producing an area strictly smaller than the remainder:

  $$\int_4^{\infty}f(x)\,dx\le R_3.$$

• Integrating from $x=3$ shifts the reference frame so the curve stays above those same rectangles, producing an area larger than the remainder:

  $$R_3\le\int_3^{\infty}f(x)\,dx.$$

Here, lower bound denotes a value smaller than $R_3$. The integral starting farther right covers less total area, which is why $\int_{N+1}^\infty f(x)\,dx$ forms the lower bound and $\int_N^\infty f(x)\,dx$ forms the upper bound.

Choosing $N$ for a target error tolerance

Let $\varepsilon$ (epsilon) be the largest error we are willing to allow. We do not need to compute $R_N$ directly; it suffices to choose an index $N$ that forces the upper bound below $\varepsilon$:

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

Because $R_N$ is bounded above by this integral, satisfying this inequality guarantees

$$0<R_N<\varepsilon.$$

Evaluating the improper integral and solving the inequality for $N$ gives a sufficient cutoff that guarantees the requested error tolerance.

Alternating series remainder estimate

An alternating series has terms that alternate in sign:

$$S=b_0-b_1+b_2-b_3+\cdots =\sum_{n=0}^{\infty}(-1)^n b_n.$$

If the positive magnitudes $b_n$ decrease monotonically toward zero,

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

the series converges by the Alternating Series Test. Successive partial sums oscillate around the true sum $S$, each step bringing the approximation closer. Consequently, the truncation error is bounded by the magnitude of the first omitted term:

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

This result is the Alternating Series Estimation Theorem.

Example: rounding to two decimal places

Consider the alternating series

$$S=\sum_{n=0}^{\infty}\frac{(-1)^n}{4^n(n+1)} =1-\frac18+\frac1{48}-\frac1{256}+\cdots.$$

Evaluating the partial sum through index $N=3$ gives

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

The first omitted term corresponds to $n=4$:

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

By the estimation theorem, the true sum $S$ lies within $b_4$ of $S_3$:

$$0.891145833\ldots \le S \le 0.892708333\ldots.$$

Because both the upper and lower bounds round to $0.89$, the estimate

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

is guaranteed correct to two decimal places.

Stopping one term earlier at $N=2$ would not provide enough precision:

$$S_2=0.895833333\ldots, \qquad |S-S_2|\le\frac1{256}=0.00390625.$$

This error interval spans from roughly $0.8919$ to $0.8997$, which contains values rounding to $0.89$ as well as values rounding to $0.90$. Thus, $N=3$ is necessary to determine the second decimal place unambiguously.

Power series

A polynomial terminates after a finite degree:

$$c_0+c_1x+c_2x^2+\cdots+c_kx^k.$$

A power series extends this concept to infinitely many terms:

$$c_0+c_1x+c_2x^2+\cdots =\sum_{n=0}^{\infty}c_nx^n.$$

More generally, a power series can be centered at any real number $a$:

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

In this expression:

• $x$ is the independent variable;
• $a$ is a constant called the center;
• $c_n$ is the coefficient of the $n$th power.

At the center $x=a$, every term with $n\ge1$ vanishes, leaving only $c_0$. Therefore, every power series converges at its center.

Radius and interval of convergence

For values of $x\ne a$, convergence depends on the distance $|x-a|$ from the center. For any power series, exactly one of three cases occurs:

1. The series converges only at the center $x=a$.
2. The series converges for all real numbers $x$.
3. There exists a real number $R>0$, called the radius of convergence, such that the series converges absolutely when $|x-a|<R$ and diverges when $|x-a|>R$:

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

When $R$ is finite, the condition $|x-a|<R$ does not specify whether convergence holds at the boundary points $x=a-R$ and $x=a+R$. Testing these endpoints individually determines the complete interval of convergence.

General procedure for finding the interval

To find the full set of values where a power series converges:

1. Identify the center $a$.
2. Apply the ratio test (or root test) to find the condition $|x-a|<R$ that ensures absolute convergence.
3. Rewrite $|x-a|<R$ as the open interval $(a-R, a+R)$.
4. Substitute each endpoint, $x=a-R$ and $x=a+R$, into the original series to obtain numerical series.
5. Test each endpoint series separately, including an endpoint in the interval if and only if its series converges.

Example: finding the interval of convergence

Find the interval of convergence for

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

The center is $a=3$. At $x=3$, every term equals $0$, so the series converges.

To determine convergence for $x\ne3$, use the Ratio Test. Recall that for a series $\sum a_n$, the Ratio Test evaluates the limit of consecutive term ratios:

• if the limit is strictly less than $1$, the series converges absolutely;
• if the limit is strictly greater than $1$, the series diverges;
• if the limit is equal to $1$, the test is inconclusive and another test must be used.

Applying this to the terms

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

the ratio becomes

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

The series converges absolutely when this limit is strictly less than $1$:

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

which simplifies to

$$2<x<4.$$

The radius of convergence is $R=1$. Now test the boundary points $x=2$ and $x=4$:

• At $x=2$, the series becomes

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

  which converges by the Alternating Series Test. Because the corresponding series of absolute values $\sum \frac{1}{n}$ diverges, this convergence is conditionally convergent. The point $x=2$ is included in the interval.

• At $x=4$, the series becomes

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

  which is the divergent harmonic series. The point $x=4$ is excluded.

Combining the interior and the valid endpoint gives the interval of convergence:

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

The bracket at $2$ denotes inclusion, while the parenthesis at $4$ denotes exclusion.

Power series from the geometric series

Many power series can be constructed directly from the geometric series formula without applying the ratio test from scratch:

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

To find a power series for a rational function, rearrange the algebraic expression into the form $\frac{A}{1-u}$, identify the expression acting as $u$, and substitute it into the summation.

Example: finding a series for $\frac{3}{2+x^2}$

Find a power series representation for

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

First, factor the denominator to match the geometric template $1-u$:

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

Here, the variable expression is

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

Substituting $u$ into the geometric expansion:

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

This geometric representation is valid whenever $|u|<1$:

$$\left|-\frac{x^2}{2}\right|<1 \quad\Longleftrightarrow\quad |x|<\sqrt2.$$

At the endpoints $x=\pm\sqrt2$, the terms $(-1)^n\frac{3}{2}$ do not approach zero, so the series diverges there. The interval of convergence is

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

Because the variable appears as $x^{2n}$, every odd power has coefficient zero. For instance, the term involving $x^6$ corresponds to $2n=6$, which is $n=3$.

Differentiating and integrating power series

Inside its open interval of convergence, a power series can be differentiated or integrated term by term, just like a polynomial.

If

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

differentiating each term gives

$$\boxed{ f'(x)=\sum_{n=1}^{\infty}n c_n(x-a)^{n-1} }.$$

Reindexing the sum so that the exponent begins at $0$ yields the equivalent form:

$$f'(x)=\sum_{n=0}^{\infty}(n+1)c_{n+1}(x-a)^n.$$

Similarly, integrating term by term introduces an arbitrary constant $C$:

$$\boxed{ \int f(x)\,dx =C+\sum_{n=0}^{\infty} \frac{c_n}{n+1}(x-a)^{n+1} }.$$

Term-by-term differentiation and integration preserve the original radius of convergence $R$. However, behavior at the endpoints $a\pm R$ can change, so endpoints must be re-evaluated after differentiating or integrating.

Differentiating the geometric series

Starting from the standard geometric series,

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

differentiating both sides with respect to $x$ gives:

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

The exponential series

Consider the series

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

Differentiating term by term:

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

The function satisfies the differential equation $f'(x)=f(x)$ with initial condition $f(0)=1$. The unique function satisfying these conditions is the natural exponential function, establishing the identity

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

for all real numbers $x$.