Module 26: Sequences, series, and the meaning of convergence

8 exercises, each with a complete solution. 1 figure. 1 note from the engine.

On this page

A. Prerequisites and observable goals
B. The central idea: infinitely many terms require a limiting definition
C. Definitions and basic tests
D. Why the tests and geometric formula hold
E. A recognition procedure
F. Worked examples
G. Independent practice
H. Complete solutions: Module 26
I. Retention and readiness

A. Prerequisites and observable goals#

Before

Limits at infinity, geometric sums, inequalities, and indexed notation.

After

Distinguish a sequence of terms from a sequence of partial sums, evaluate geometric and telescoping series, and use the term test only in its valid direction.

B. The central idea: infinitely many terms require a limiting definition#

A sequence is a function whose inputs are successive integers: $a_1,a_2,\ldots$ . A series is an instruction to add sequence terms, interpreted through the partial sums $s_N=\sum_{n=1}^N a_n$ . The statement $\sum_{n=1}^\infty a_n=S$ means $s_N\to S$ , not that an infinite addition can be treated as an ordinary finite sum without justification. The limit of $a_n$ and the limit of $s_N$ answer different questions.

A series is the limit of its partial sums. Left: a unit square cut into pieces of area

\tfrac12,\tfrac14,\tfrac18,\ldots

; the pieces never overflow the square and leave less and less of it uncovered. Right: the partial sums

s_N=1-1/2^N

climb toward

1

. Saying that the series equals

1

means exactly that

s_N\to1

C. Definitions and basic tests#

A sequence converges to $L$ if for every $\varepsilon>0$ there is an integer $N$ such that $n\ge N$ implies $|a_n-L|<\varepsilon$ . This is the sequence version of a limit: the input eventually exceeds every fixed cutoff, rather than moving continuously near a point.

If a series converges, then $a_n\to0$ . Consequently, if the terms fail to tend to zero, the series diverges. The converse is false: $a_n\to0$ is necessary but not sufficient. A monotone bounded sequence converges; the bounds and monotonicity must apply to the sequence under discussion, whether terms or partial sums.

For a geometric series with first term $A$ and ratio $r$ ,

\sum_{n=0}^{\infty}Ar^n=\frac{A}{1-r}\quad\text{when }|r|<1.

If $A\ne0$ and $|r|\ge1$ , it diverges. The exceptional zero sequence when $A=0$ must not be excluded by an overgeneralized slogan. Convergent series can be added or multiplied by constants; arbitrary rearrangement of conditionally convergent series is a later caution, not a generally free operation. [O2Seq]

D. Why the tests and geometric formula hold#

If $s_N\to S$ , then $a_N=s_N-s_{N-1}\to S-S=0$ , proving the necessary term condition. Notice why the proof cannot be reversed: small increments need not prevent an accumulated total from growing indefinitely.

For a finite geometric sum $S_N=A+Ar+\cdots+Ar^{N-1}$ , subtract $rS_N$ to get $(1-r)S_N=A(1-r^N)$ . For $r\ne1$ ,

S_N=\frac{A(1-r^N)}{1-r}.

Only after this finite identity is established do we take the limit. If $|r|<1$ , $r^N\to0$ , giving the infinite formula. The indexing determines how many powers occur; check the first term before applying it.

E. A recognition procedure#

For a sequence, compute or bound its limiting terms. For a series, first inspect whether terms can tend to zero. Then look for a constant ratio (geometric) or cancellation in a finite partial sum (telescoping). Write several terms or the partial sum explicitly rather than canceling imagined infinities. If no special pattern applies, use the general tests in the next modules. A convergence test usually does not reveal an exact sum value.

F. Worked examples#

A sequence limit. For $a_n=(3n^2+1)/(2n^2-5)$ , divide numerator and denominator by $n^2$ to find $a_n\to3/2$ . Therefore a series having these terms diverges by the term test, even though the sequence itself converges.

A telescoping sum. Since $1/[n(n+1)]=1/n-1/(n+1)$ ,

\sum_{n=1}^N\frac1{n(n+1)}=1-\frac1{N+1}\longrightarrow1.

The cancellation is performed in a finite sum. The exact tail after $N$ terms is $1/(N+1)$ , information stronger than mere convergence.

Why term decay is insufficient. In the harmonic series, terms from $2^{k-1}+1$ through $2^k$ contribute at least $2^{k-1}/2^k=1/2$ . There are arbitrarily many such blocks, so partial sums grow without bound even though $1/n\to0$ . Grouping these positive terms is safe and provides a direct divergence proof.

From the engine: ZeroVectorA geometric series in every texture

A texture in ZeroVector is stored with a chain of smaller copies, each half the width and height of the one before, down to a single pixel, so a distant surface can be drawn from a copy near its size on screen. Each level has a quarter of the pixels of the one before, so the memory of the whole chain is a geometric series with ratio $1/4$ .

ZeroVector: compressed texture sizesC++

// Bytes of one level: whole blocks, so a 1254-wide BC7 level is 314 blocks
// across. 0 for an unsupported format or a zero size.
inline u64 ktx2LevelByteSize(u32 vkFormat, u32 width, u32 height) {
    const Ktx2BlockLayout layout = ktx2BlockLayout(vkFormat);
    if (layout.blockSide == 0 || width == 0 || height == 0) return 0;
    const u64 blocksWide = (u64{width} + layout.blockSide - 1) / layout.blockSide;
    const u64 blocksHigh = (u64{height} + layout.blockSide - 1) / layout.blockSide;
    return blocksWide * blocksHigh * layout.bytesPerBlock;
}

// Levels of a full chain down to 1x1, as Vulkan and the engine count them.
inline u32 ktx2FullMipCount(u32 width, u32 height) {
    u32 side = std::max(width, height);
    u32 levels = 1;
    while (side > 1) {
        side >>= 1;
        ++levels;
    }
    return levels;
}

The first function counts whole blocks, because BC7, the format its comment names, stores 16 bytes for each 4-by-4 block of pixels: adding 3 before the integer division by 4 rounds up, so a level 1254 pixels wide is 314 blocks across. The second counts levels: for $2048\times 2048$ there are 12. Level 0 takes $512\cdot 512\cdot 16 = 4{,}194{,}304$ bytes, a quarter of the 16,777,216 the same image needs uncompressed at four bytes a pixel, and the full chain costs $1 + \tfrac14 + \tfrac1{16} + \cdots$ times that, which this module sums to $\tfrac{1}{1 - 1/4} = \tfrac43$ . The exact total is a shade more, because the two smallest levels still occupy one whole block each.

G. Independent practice#

F foundation C standard course work T transfer S optional stretch L runnable lab or project. Attempt the complete set before opening its solutions.

M26.01FSolution

Find the limit of $a_n=(5n-2)/(2n+7)$ . Does $\sum_{n=1}^{\infty}a_n$ converge?

M26.02CSolution

Determine whether $a_n=(-1)^n$ and $b_n=(-1)^n/n$ converge as sequences.

M26.03CSolution

Evaluate $\sum_{n=0}^{\infty}3(2/5)^n$ . Find the exact error after terms $n=0$ through $N-1$ .

M26.04TSolution

Evaluate $\sum_{n=2}^{\infty}(-1/3)^n$ .

M26.05CSolution

Evaluate $\sum_{n=1}^{\infty}2/[n(n+2)]$ using a finite partial sum.

M26.06TSolution

Explain why knowing $a_n\to0$ does not determine whether $\sum a_n$ converges. Give two positive examples with opposite outcomes.

M26.07TSolution

Let $a_1=1$ and $a_{n+1}=\sqrt{2+a_n}$ . Prove the sequence converges and find its limit.

M26.08SSolution

A ball is dropped from height $5$ m and rebounds to $3/5$ of the previous height each time. Under this idealized model, find the total vertical distance traveled through infinitely many rebounds.

Optional small hints

For M26.05, decompose into $1/n-1/(n+2)$ and identify the two uncanceled terms at each end. For M26.07, prove both an upper bound and monotonicity before passing to the recurrence limit. For M26.08, the initial fall is counted once; later rebound heights are traversed twice.

H. Complete solutions: Module 26#

Compare the first step where your work differs. A corrected answer is not yet an independent success.

M26.01Show solution

A convergent term sequence can produce a divergent series

Dividing by $n$ gives $a_n=(5-2/n)/(2+7/n)\to\boxed{5/2}$ . Since that nonzero limit violates the necessary condition for a convergent series, $\boxed{\sum a_n\text{ diverges}}$ . The sequence convergence is not the series convergence.

Back to M26.01

M26.02Show solution

Oscillation may or may not shrink

The even and odd subsequences of $(-1)^n$ have limits $1$ and $-1$ , so no common limit exists. But $|(-1)^n/n|=1/n\to0$ , so squeezing gives $\boxed{b_n\to0}$ . Oscillation alone is not a divergence test for sequences; its amplitude matters.

Back to M26.02

M26.03Show solution

A geometric value and its exact tail

The ratio is $2/5$ , of magnitude less than one. Hence the sum is $3/(1-2/5)=\boxed5$ . After the first $N$ terms the partial sum is $5[1-(2/5)^N]$ , so the exact positive remainder is $\boxed{5(2/5)^N}$ . The error is indexed by the number of included terms, not by an unexplained last exponent.

Back to M26.03

M26.04Show solution

Use the actual first term

The first included term is $(-1/3)^2=1/9$ , with ratio $-1/3$ . Thus the sum is $(1/9)/(1+1/3)=\boxed{1/12}$ . Starting at $n=0$ instead would give a different total; the first-term check catches that indexing error.

Back to M26.04

M26.05Show solution

Two boundary terms survive at each end

Since $2/[n(n+2)]=1/n-1/(n+2)$ , the partial sum is

S_N=1+\frac12-\frac1{N+1}-\frac1{N+2}.

Taking $N\to\infty$ gives $\boxed{3/2}$ . All cancellation took place in the finite identity, so no subtraction of divergent infinite harmonic series was used.

Back to M26.05

M26.06Show solution

A necessary condition cannot decide the outcome

Both $a_n=1/n$ and $b_n=1/[n(n+1)]$ are positive and tend to zero. The first has divergent partial sums by the block comparison explained above; the second telescopes to a finite sum $1$ . Thus the zero term limit is compatible with either outcome and cannot certify convergence.

Back to M26.06

M26.07Show solution

Bounded monotonicity justifies taking a limit

The base term satisfies $1\le a_1\le2$ . If $1\le a_n\le2$ , then $1\le\sqrt{2+a_n}\le2$ , establishing the bound by induction. Also for $1\le a_n\le2$ , $2+a_n\ge a_n^2$ because $(2-a_n)(a_n+1)\ge0$ , so $a_{n+1}\ge a_n$ . The sequence is increasing and bounded, hence converges to some $L\in[1,2]$ . Continuity of square root gives $L=\sqrt{2+L}$ , so $(L-2)(L+1)=0$ . Only $\boxed{L=2}$ lies in the established interval.

Back to M26.07

M26.08Show solution

Translate repeated motion into a geometric sum

The initial drop contributes $5$ m. Rebound heights are $5(3/5)^n$ for $n\ge1$ , each traveled upward and downward. Thus

D=5+2\sum_{n=1}^{\infty}5(3/5)^n =5+10\frac{3/5}{1-3/5}=\boxed{20\text{ m}}.

This is a statement about the idealized distance model; it does not assert that real impacts continue indefinitely or that the same simple law describes their timing.

Back to M26.08

I. Retention and readiness#

Retrieve M16.05 and M24.01. Explain which sequence must converge to define a series sum. Advance when you can reject “the terms go to zero, therefore it converges” and can keep starting indices straight in a new geometric problem.