Sum of infinite GP — when it converges, proof: Question bank

True or false — justify

Statement is "the infinite GP converges" or a claim about it — decide, then give the reason.

A GP with

r = - \frac{1}{2}

converges.

True. Convergence depends on

∣ r ∣ = \frac{1}{2} < 1

, and the sign of

r

is irrelevant to shrinking — the terms still get smaller in size, just alternating.

If every term of an infinite GP is positive, the series must converge.

False.

1 + 2 + 4 + \dots

has all-positive terms but

r = 2

, so terms grow and it diverges. Positivity says nothing about

∣ r ∣

A GP with

r = 1

can converge if

a

is small enough, like

a = 0.0001

False. With

r = 1

every term equals

a

, so the sum is

a + a + a + \dots

, which grows without bound for any

a \neq = 0

no matter how tiny.

If the terms of an infinite GP tend to

0

, the sum is guaranteed finite.

True for a GP specifically: terms

\to 0

forces

∣ r ∣ < 1

, which is exactly the convergence condition. (This shortcut is special to GPs — it fails for general series like

\sum 1/ n

0. \overline{9}

(i.e.

0.999 \dots

) as a GP sum equals

1

exactly, not "just below

1

True. It is

a = 0.9, r = 0.1

, giving

\frac{0.9}{1 - 0.1} = \frac{0.9}{0.9} = 1

. The infinite sum is the limit, and that limit is exactly

1

A GP with

r = - 1

has partial sums that approach

\frac{a}{2}

, so it converges to

\frac{a}{2}

False. Its partial sums are

a, 0, a, 0, \dots

— they oscillate forever and never approach a single value, so no limit (and no sum) exists.

S_{\infty}

is negative, the GP must have been divergent.

False. A convergent GP can have a genuine negative sum, e.g.

a = - 2, r = \frac{1}{2}

gives

S_{\infty} = \frac{- 2}{1/2} = - 4

. Sign of the answer says nothing about convergence.

Doubling the first term

a

doubles

S_{\infty}

True (when it converges).

S_{\infty} = \frac{a}{1 - r}

is directly proportional to

a

, so scaling

a

by any factor scales the whole sum by that same factor;

r

is untouched.

Spot the error

Each shows a plausible-looking solution. Name the mistake.

1 + 2 + 4 + 8 + \dots = \frac{1}{1 - 2} = - 1

The formula

\frac{a}{1 - r}

was applied with

r = 2

, but it is only valid for

∣ r ∣ < 1

. Here

r^{N}

blows up instead of vanishing, so the closed form is meaningless — the series actually diverges.

"For

6 + 2 + \frac{2}{3} + \dots

, we have

a = 2

and

r = 3

Two errors:

a

is the first term

6

, not

2

; and

r = \frac{next}{current} = \frac{2}{6} = \frac{1}{3}

, not its reciprocal

3

. The correct sum is

\frac{6}{1 - 1/3} = 9

3 - 1 + \frac{1}{3} - \dots

diverges because some terms are negative."

Negativity is fine; what matters is

∣ r ∣ = - \frac{1}{3} = \frac{1}{3} < 1

, so it converges (to

\frac{9}{4}

). Alternating signs do not cause divergence.

0. \overline{7} = 0.7 + 0.07 + \dots

, and since it never stops adding, its sum is infinite."

The terms shrink by

r = 0.1

each time (

∣ r ∣ < 1

), so the growing tail contributes ever-tinier amounts and the total settles on the finite value

\frac{0.7}{0.9} = \frac{7}{9}

"To get

r

I divide the first term by the second:

r = \frac{a _{1}}{a _{2}}

Upside-down. The ratio is later over earlier,

r = \frac{a _{2}}{a _{1}}

. Inverting it will falsely report a shrinking GP as growing (and vice versa).

"The partial-sum formula

S_{N} = \frac{a ( 1 - r ^{N} )}{1 - r}

works for every

r

, so I'll use it with

r = 1

r = 1

we divided by

1 - r = 0

during the derivation, so the formula is undefined there. For

r = 1

the sum is simply

S_{N} = N a

(add

a

to itself

N

times).

"Since the formula gives a number for

r = 1.5

, the series must converge."

Getting a number is not the same as converging. The algebra of

\frac{a}{1 - r}

runs regardless, but the derivation that justifies it required

r^{N} \to 0

, which fails for

∣ r ∣ \geq 1

. The number is a fake.

Why questions

Explain the mechanism, not just the rule.

Why must

∣ r ∣ < 1

rather than just

r < 1

Because convergence hinges on the size

∣ r^{N} ∣

shrinking to

0

. If

r = - 2

then

r < 1

is true but

∣ r ∣ = 2

, so

∣ r^{N} ∣

grows — the "

< 1

" without absolute value would wrongly admit diverging cases.

Why does the whole convergence question reduce to the single term

r^{N}

S_{N} = \frac{a ( 1 - r ^{N} )}{1 - r}

, the parts

a

and

1 - r

are fixed constants;

r^{N}

is the only piece that changes as

N

grows, so its behaviour alone decides whether

S_{N}

settles.

Why does the "shift-and-subtract" trick make the middle terms disappear?

Multiplying

S_{N}

r

shifts every term one slot to the right, so each interior term now sits under an identical copy of itself; subtracting cancels them, leaving only the unmatched first term

a

and the leaked-out last term

a r^{N}

Why can an infinite number of positive bites still add to a finite total?

Because with

∣ r ∣ < 1

each bite is a fixed fraction of the previous one, so the remaining amount shrinks geometrically; the leftovers approach

0

, and "total minus leftover" approaches a fixed number. (See Limits of Sequences for

r^{N} \to 0

Why is

0. \overline{3} = \frac{1}{3}

not an approximation but an exact equality?

The notation

0. \overline{3}

means the limit of the partial sums

0.3, 0.33, 0.333, \dots

, and that limit is exactly

\frac{0.3}{1 - 0.1} = \frac{1}{3}

— see Recurring Decimals as Fractions.

Why does a perpetual payment stream (a perpetuity) have a finite present value?

Each future payment is discounted by a factor

r < 1

, forming a convergent GP; distant payments are worth vanishingly little today, so the infinite stream sums to a finite value — see Present Value & Perpetuities.

Why is checking

∣ r ∣ < 1

a precondition, not an afterthought?

The formula

\frac{a}{1 - r}

was derived by assuming

r^{N} \to 0

. Using it first and checking later means you may have already trusted a value that the assumption never permitted — the check must gate the formula, not follow it.

Edge cases

The boundaries that break naive intuition.

r = 0

: what is the sum

a + 0 + 0 + \dots

It converges trivially to

a

, and the formula agrees:

\frac{a}{1 - 0} = a

. (Strictly

0^{0}

in the first term is read as the term

a

itself.)

a = 0

with any

r

: does it converge?

Yes — every term is

0

, so

S_{\infty} = 0

. The formula gives

\frac{0}{1 - r} = 0

for any

r \neq = 1

; the ratio never matters when there is nothing to add.

r

exactly

1

: converge or diverge?

Diverges for

a \neq = 0

. Every term equals

a

, so

S_{N} = N a \to \pm \infty

. This is the case the formula literally cannot touch (division by

1 - r = 0

r

exactly

- 1

: converge or diverge?

Diverges (no sum). Partial sums oscillate

a, 0, a, 0, \dots

and never approach one number — oscillation is a form of divergence, not a compromise value.

∣ r ∣

just barely less than

1

, like

r = 0.9999

: still converges?

Yes. Any

∣ r ∣ < 1

, however close to

1

, forces

r^{N} \to 0

eventually — it just converges slowly. Convergence is a yes/no matter of the boundary, not of how close you are to it.

∣ r ∣

just barely more than

1

, like

r = 1.0001

: converge?

No. Even a hair above

1

makes

∣ r^{N} ∣

grow without bound (slowly at first, then explosively), so the series diverges. The boundary

∣ r ∣ = 1

is a hard cliff.

A GP where terms shrink but never reach zero — is "never reaching zero" a problem?

No. Convergence needs the terms to approach

0

, not equal it. With

∣ r ∣ < 1

each term stays nonzero yet gets arbitrarily small, which is exactly enough for the total to settle.

Connections

Parent: Sum of infinite GP
Geometric Progression — nth term
Sum of finite GP
Limits of Sequences
Convergence of Series
Recurring Decimals as Fractions
Present Value & Perpetuities