Series and Convergence

Sums of Squares and Cubes #

Closed formulas for the squares and the cubes.

The first n squares and the first n cubes each add to a closed formula

The whole numbers from 1 to n add to n(n + 1)/2.

The squares have their own formula, and n = 3 checks it: both sides come to 14.

The cubes add to the square of Σr: 1 + 8 + 27 is 36, which is 6², and 1 + 2 + 3 is 6.

A sum splits term by term, so a mixed expression is built from the three formulas.

Now you

Σ(r² + 2r) splits into

Σr³ for r = 1 to 3 equals

The Method of Differences #

Terms that cancel in pairs collapse a sum.

When each term is a difference of consecutive values, everything inside cancels and only the ends survive

Split each term into a difference of two consecutive pieces.

Written out, every piece meets its opposite in the next bracket and cancels.

The first piece and the last piece survive, so the sum is 1 − 1/(n + 1).

As n grows, 1/(n + 1) shrinks toward 0 and the total approaches exactly 1.

Now you

As n grows without bound, Σ 1/(r(r + 1)) tends to

1/(1·2) + 1/(2·3) + 1/(3·4) equals

The nth Term Test for Divergence #

Terms that refuse to die rule a sum out.

A series whose terms do not shrink to zero cannot converge, though shrinking terms guarantee nothing

If the partial sums converge, each new term is the gap between two partial sums that are both close to the limit.

Turn it around: the terms of n/(2n + 1) tend to 1/2, so that series cannot converge.

The converse is false: 1/n does shrink to 0 and its sum still grows past every bound.

So this test can only ever say diverges — never that a series converges.

Now you

Terms shrinking to zero proves

What does the nth term test say about Σ n/(2n + 1)?

The p-Series #

One over n to the p settles only past 1.

The sum of 1 over n to the power p converges exactly when p is greater than 1

The p-series Σ1/nᵖ is one family, with the harmonic series at p = 1.

Each term 1/n is a strip one wide, standing on the curve y = 1/x at its left edge.

The strips and the area under the curve bound each other, so the sum and the integral converge or diverge together.

That integral is finite out to infinity exactly when p is greater than 1.

So Σ1/n² converges, while Σ1/n and Σ1/√n both diverge.

Now you

Σ 1/nᵖ converges exactly when

Does Σ 1/√n converge or diverge?

Comparison Tests for Series #

Judge a series by a simpler one beside it.

A positive series is judged by a simpler one it stays below or keeps pace with

A positive series whose terms stay below a convergent series must converge too.

And a series whose terms stay above a divergent series must diverge too.

So Σ1/(n² + 1) converges: every term sits below the matching term of Σ1/n².

Sometimes the inequality points the wrong way, and direct comparison gives no answer.

Take the ratio instead: if it tends to a finite non-zero limit, both series behave alike.

Against 1/n² the ratio tends to 2, so Σ(2n + 1)/(n³ + 5) converges as well.

Now you

Σ (2n + 1)/(n³ + 5) is best compared with

aₙ/bₙ → 3 and Σbₙ diverges, so Σaₙ

The Alternating Series Test #

Shrinking terms with flipping signs settle.

Alternating terms shrinking to zero close on the sum from both sides at once

Partial sums of 1 − 1/2 + 1/3 − 1/4 + … jump over the limit, then back under it.

The conditions: the signs alternate, the sizes fall, and the sizes reach 0.

Each step overshoots by less than the last, so the limit is trapped between consecutive totals.

So the error from stopping early is less than the first term you leave out.

Stopping after four terms, the limit is within 1/5 of the partial sum shown.

Drop the signs and it diverges, so this series is only conditionally convergent.

Now you

The alternating series test needs the sizes to

The series 1 − 1/2 + 1/3 − 1/4 + … is

The Ratio Test #

Each term against the last decides the sum.

The limit of each term divided by the one before decides whether a series converges, unless that limit is exactly one

A geometric series converges when each term is a fixed fraction of the one before. The ratio test applies the same idea to any series.

Let L be the limit of each term divided by the one before: L < 1 converges, L > 1 diverges.

For Σ2ⁿ/n! the ratio is 2/(n + 1), which tends to 0, so the series converges.

For Σn!/2ⁿ, the same terms inverted, the ratio is (n + 1)/2, which grows past 1, so that series diverges.

Every p-series gives L = 1, yet some converge and some do not, so L = 1 decides nothing.

Now you

By the ratio test, Σ 2ⁿ/n!

When L = 1, the ratio test

Radius and Interval of Convergence #

Where a power series settles, ends included.

The ratio test gives a power series a radius, and the two endpoints are then tested one at a time

A power series contains x, so whether it converges depends on which x you put in.

The ratio test needs L < 1, and that says x must lie within R of the center.

That is an interval: the center, and a distance R either side of it.

For Σxⁿ/n the ratio tends to |x|, so the radius is 1 and the center is 0.

At x = 1 it is the harmonic series, which diverges; at x = −1 it alternates and converges.

So the interval is −1 up to but not including 1. Always test the two endpoints separately.

Now you

The radius of convergence of Σ xⁿ/n is

The radius of convergence of Σ xⁿ/n! is

The Standard Maclaurin Series #

Five series that generate the rest.

The geometric series generates the logarithm series and sine generates cosine, so few need building from scratch

Start from a series already known: 1/(1 − x) is the geometric sum, for |x| < 1.

Replace x by −x: the series becomes 1/(1 + x), with alternating signs.

Integrate both sides term by term and the series for ln(1 + x) appears.

Differentiating the sine series term by term gives the cosine series.

These five series give the rest: substitute, differentiate or integrate to reach a new one.

Now you

The Maclaurin series of ln(1 + x) begins

1/(1 − x) expands as

Taylor Series About a Point #

Derivatives read at a center other than zero.

Measuring the derivatives at a instead of at zero rebuilds a function near a

ln x has no value at x = 0, so nothing can be measured there. Build the series at x = 1.

The formula is the same with a new center: derivatives at a, and powers of x − a.

At x = 1, ln x and its derivatives take the values 0, 1, −1, 2; divide each by n! to get the coefficients.

Two terms already follow ln x closely near 1, and drift away as x moves further off.

Setting a = 0 returns the Maclaurin series, so the center is simply a choice.

Now you

The Taylor series of ln x about a = 1 begins

Term n of a Taylor series about a is

The Lagrange Error Bound #

How far a Taylor polynomial can be off.

The error left by a Taylor polynomial is at most the next term, with its derivative replaced by an upper bound

Cutting a Taylor series after degree n leaves a remainder — how big can it be?

The error is at most M times the next power over the next factorial.

So the bound is the next term, with the unknown derivative replaced by its largest size.

Take sin x to degree 3 on |x| ≤ 1: every derivative of sine has size at most 1.

The bound is 1/24, and the true error is smaller — a bound is the most the error can be, not the error itself.

Now you

Approximating sin x by its degree-3 polynomial on |x| ≤ 1, the error is at most

In the Lagrange bound, M stands for