Series — partial sums, convergence definition: Question bank

Before we start, three words you must have straight (each is built fully in the parent note):

term $a_{n}$ — the individual number at position $n$ in the list you are adding.
partial sum $s_{N} = a_{1} + a_{2} + \dots + a_{N}$ — the running total after $N$ terms.
the series converges — means the sequence of running totals $(s_{N})$ has a finite limit.

Keep the three separate: terms, running totals, and the limit of the running totals. Almost every trap below is one of these three quietly swapped for another.

True or false — justify

A term list

\to 0

guarantees convergence

False — the Harmonic series

\sum 1/ n

has

a_{n} \to 0

yet its running totals grow without bound. Small terms can still accumulate to infinity if they shrink too slowly.

If a series converges then its terms must

\to 0

True — since

a_{N} = s_{N} - s_{N - 1}

and both

s_{N}, s_{N - 1} \to S

, the terms tend to

S - S = 0

. This is the n-th Term Test's honest direction.

"Diverges" always means the running total marches off to

+ \infty

False —

\sum (- 1)^{n}

has partial sums

- 1, 0, - 1, 0, \dots

which stay bounded but never settle on one number, so it diverges without going to infinity.

\sum a r^{n - 1} = \frac{a}{1 - r}

holds for every

r \neq = 1

False — the formula is only valid when

∣ r ∣ < 1

, because it was born from

r^{N} \to 0

, which fails once

∣ r ∣ \geq 1

The sequence

(a_{n})

and the series

\sum a_{n}

are basically the same object

False —

(a_{n})

is the list of terms;

\sum a_{n}

lives or dies by the separate sequence of running totals

(s_{N})

. They can behave oppositely (e.g.

a_{n} \to 0

but

\sum a_{n}

diverges).

Removing or changing the first ten terms can turn a divergent series convergent

False — a finite change only shifts every later partial sum by a fixed constant, which cannot create or destroy a limit. Convergence is a "tail" property.

s_{N}

is a bounded sequence, the series converges

False — bounded is not the same as convergent;

\sum (- 1)^{n}

has bounded partial sums (

- 1, 0

) that never approach a single value.

A telescoping series always converges

False — telescoping just gives a clean closed form for

s_{N}

; whether it converges still depends on whether that closed form has a finite limit. See Telescoping series.

If two series both diverge, their sum

\sum (a_{n} + b_{n})

must diverge

False — take

a_{n} = 1

and

b_{n} = - 1

; each diverges but the combined series has all-zero terms and converges to

0

Spot the error

a_{n} = 1/ n \to 0

, therefore

\sum 1/ n

converges." Where's the flaw?

The n-th Term Test only runs one way: convergence forces

a_{n} \to 0

, but

a_{n} \to 0

does not force convergence. The harmonic series is the standing counterexample.

\sum 2 \cdot 3^{n - 1} = \frac{2}{1 - 3} = - 1

." What went wrong?

The geometric formula was applied with

r = 3

, but

∣ r ∣ \geq 1

, so the series actually diverges — there is no finite sum, and

- 1

is a meaningless plug-in.

"The series equals the sum

a_{1} + a_{2} + \dots

, so I'll just add them all up." Why is this not a definition?

Addition is a binary operation — you can only ever add two numbers at a time — so "add infinitely many" is undefined. The real definition routes through

lim_{N \to \infty} s_{N}

s_{N} \to S

means every partial sum equals

S

from some point on." Correct?

No — the

ε

-definition says every partial sum eventually lands within a band of half-width

ε

S

, not exactly on

S

. They approach, they need never arrive.

"Since each block of the harmonic series is a bit more than

1/2

and there are infinitely many blocks, the sum is

\infty

." Is the logic valid?

Yes — adding more than

1/2

infinitely often makes

s_{N}

exceed any bound, which is exactly a divergent (unbounded) running total. This one is correct reasoning, the trap being to distrust it.

\sum (- 1)^{n}

diverges because its terms don't go to

0

." Right conclusion, but check the reason.

The conclusion is right and the reason is right:

(- 1)^{n}

oscillates

- 1, 1, \dots

and never approaches

0

, so the n-th Term Test kills it. The trap is realising a bounded series can still fail the term test.

Why questions

Why do we define convergence through partial sums instead of the raw infinite sum?

Because "add forever" is not a legal operation, whereas a sequence of running totals either has a limit or doesn't — converting an impossible task into one we already solved with Sequences — limits and convergence.

Why does multiplying

s_{N}

r

and subtracting collapse the geometric partial sum?

The factor

r

shifts every term one slot to the right, so all the middle terms line up and cancel, leaving only the first term

a

and the last leftover

- a r^{N}

— a telescoping cancellation.

Why is

∣ r ∣ < 1

the exact border for geometric convergence?

Because the only

N

-dependent piece of

s_{N}

r^{N}

, and

r^{N} \to 0

precisely when

∣ r ∣ < 1

; at

∣ r ∣ = 1

it fails to shrink, and at

∣ r ∣ > 1

it blows up.

Why is the n-th Term Test only a test for divergence and never for convergence?

Its only guarantee runs contrapositively —

a_{n} \neq \to 0

proves divergence — but

a_{n} \to 0

leaves the outcome open, so it can never certify convergence.

Why can't a single very large but finite partial sum tell us the series diverges?

Convergence is about the tail behaviour of

(s_{N})

N \to \infty

; any finite

s_{N}

is just one snapshot, and the running total could still settle far beyond it.

Why does the parent note insist "always go through

s_{N}

Because the two most common errors both come from reasoning about the terms

(a_{n})

directly, when the series' fate is decided entirely by the limit of the running totals

(s_{N})

Edge cases

What happens to

\sum a r^{n - 1}

when

r = 1

Every term equals

a

, so

s_{N} = a N

, which marches to

\pm \infty

(for

a \neq = 0

) — the series diverges, which is why the formula excludes

r = 1

What happens to

\sum a r^{n - 1}

when

r = - 1

The partial sums bounce

a, 0, a, 0, \dots

and never settle, so it diverges — bounded oscillation, not a blow-up.

What is the sum when

a = 0

(every term zero)?

All partial sums are

0

, so the series trivially converges to

0

regardless of

r

— a degenerate but valid convergent case.

Does the harmonic series ever "stop" growing since the terms become microscopic?

No — although

1/ n \to 0

, the partial sums pass every finite bound eventually; the terms shrink but not fast enough to cap the total.

Can a series converge if infinitely many terms are exactly

0

Yes — zero terms add nothing to the running total, so convergence depends only on the nonzero terms; e.g.

\sum

1, 0, \frac{1}{4}, 0, \frac{1}{9}, \dots

behaves like the surviving p-series.

lim_{N \to \infty} s_{N} = + \infty

, is that "a limit"?

In the convergence sense, no — a series converges only to a finite number, so an infinite limit counts as divergence even though

s_{N}

has a definite (infinite) trend.

Connections

Parent: partial sums & convergence
n-th Term Test for Divergence — the source of the biggest trap here
Harmonic series — the eternal counterexample to " $a_{n} \to 0 \Rightarrow$ converges"
Geometric series — where the $∣ r ∣ < 1$ passport traps live
Telescoping series — clean $s_{N}$ does not guarantee a limit
Sequences — limits and convergence — every trap reduces to a sequence limit