Sigma (Σ) notation — evaluating, telescoping sums: Question bank

The reveals below use the Question ::: Answer format — the part after ::: is the worked justification, never a bare "yes/no".

Quick visual refresher (so this page stands alone)

Before the traps, here is every piece of machinery they attack — with a picture, so you never have to chase the parent note mid-question.

Figure — Sigma (Σ) notation — evaluating, telescoping sums

True or false — justify

\sum_{k = 1}^{n} (a_{k} b_{k}) = (\sum_{k = 1}^{n} a_{k}) (\sum_{k = 1}^{n} b_{k})

False. Sigma is linear over

+

and constant multiples only; products of terms do not distribute. Test

\sum_{k = 1}^{2} k \cdot k = 5

but

(\sum k)^{2} = 9

The name of the index in

\sum_{k = 1}^{n} k^{2}

matters — writing

\sum_{j = 1}^{n} j^{2}

gives a different value.

False. The index is a dummy variable — it is used up ticking through

1, \dots, n

inside the machine and never appears in the answer, so

k

j

i

all give the identical closed form

\frac{n ( n + 1 ) ( 2 n + 1 )}{6}

\sum_{k = 1}^{n} c = c

for any constant

c

False. The constant

c

is added once for each of the

n

terms, so the total is

n c

, not

c

. Only when

n = 1

do they coincide.

Every fraction of the form

\frac{1}{k ( k + 1 )}

telescopes because it has

k

and

k + 1

in the denominator.

True in effect, but the reason is precise: partial fractions split it into

\frac{1}{k} - \frac{1}{k + 1}

, which is exactly the shape

f (k) - f (k + 1)

, and only that shape telescopes — the factored denominator is just the clue.

a_{k} = f (k) - f (k + 1)

then

\sum_{k = 1}^{n} a_{k}

always equals

f (1) - f (n + 1)

True. Each term's tail

- f (k + 1)

is cancelled by the next term's head

+ f (k + 1)

; only the very first head

f (1)

and very last tail

- f (n + 1)

have no partner and survive.

\sum_{k = 1}^{n} k^{3} = (\sum_{k = 1}^{n} k)^{2}

True — a genuine identity, both equal

[\frac{n ( n + 1 )}{2}]^{2}

. But do not generalise:

\sum k^{2} \neq = (\sum k)^{2/1}

or anything similar; this coincidence is special to cubes.

Reversing the order of terms in a finite sum can change its value.

False. Addition of finitely many numbers is commutative and associative, so any reordering (including Gauss's forwards+backwards trick) preserves the total.

Spot the error

A student rewrites

\sum_{k = 1}^{n} \frac{1}{k + 1}

\sum_{j = 2}^{n + 1} \frac{1}{j}

but keeps the limits as

1

n

. Find the mistake.

The substitution

j = k + 1

moves both limits:

k = 1 \Rightarrow j = 2

and

k = n \Rightarrow j = n + 1

. Keeping

1

n

leaves the wrong bounds — every index shift must carry the limits along with it.

A student claims

\sum_{k = 1}^{n} \frac{1}{k ( k + 2 )}

telescopes to

\frac{1}{1} - \frac{1}{n + 2}

, only two survivors. Where's the slip?

First get the split honestly:

\frac{1}{k ( k + 2 )} = \frac{A}{k} + \frac{B}{k + 2}

gives

1 = A (k + 2) + B k

;

k = 0 \Rightarrow A = \frac{1}{2}

k = - 2 \Rightarrow B = - \frac{1}{2}

, so it is

\frac{1}{2} (\frac{1}{k} - \frac{1}{k + 2})

. The gap is

2

, so

\frac{1}{k}

cancels

\frac{1}{k + 2}

two rows later, not one. Two heads

(\frac{1}{1}, \frac{1}{2})

and two tails

(- \frac{1}{n + 1}, - \frac{1}{n + 2})

survive, all carrying that

\frac{1}{2}

factor.

Someone writes

\sum_{k = 1}^{n} (2 k + 3) = 2 \sum k + 3

. What went wrong?

The constant

3

is inside the sum, so it is added

n

times:

\sum 3 = 3 n

, not

3

. Correct answer:

2 \cdot \frac{n ( n + 1 )}{2} + 3 n

A student factors

\frac{1}{k ( k + 1 )} = \frac{A}{k} + \frac{B}{k + 1}

and gets

A = 1, B = 1

. Diagnose.

Sign error. Clearing denominators gives

1 = A (k + 1) + B k

; setting

k = 0 \Rightarrow A = 1

, setting

k = - 1 \Rightarrow B = - 1

. It must be

\frac{1}{k} - \frac{1}{k + 1}

— a difference, or nothing telescopes.

A student says

\sum_{k = 3}^{7} 1

has

7 - 3 = 4

terms. Correct it.

Off by one. The count is

n - m + 1 = 7 - 3 + 1 = 5

(namely

k = 3, 4, 5, 6, 7

). Subtracting endpoints always loses one, so add it back.

A student concludes

\sum_{k = 1}^{n} [f (k) - f (k + 1)] = f (1) - f (n)

. Fix the endpoint.

The last term is

f (n) - f (n + 1)

, so the surviving tail is

- f (n + 1)

, not

- f (n)

. The correct result is

f (1) - f (n + 1)

A student telescopes

\sum_{k = 1}^{n} [f (k + 1) - f (k)]

and writes

f (1) - f (n + 1)

. Spot the reversal.

This form is

+ f (k + 1) - f (k)

, the opposite orientation, so it collapses to

f (n + 1) - f (1)

. Direction of the difference flips the surviving endpoints — always check which piece is the head.

Why questions

Why does the "

+ 1

" appear in the term count

n - m + 1

Because

n - m

counts the gaps between labels, and there is one more label than gaps (a fence with

n - m

gaps has

n - m + 1

posts).

Why do we use partial fractions to attack

\sum \frac{1}{k ( k + 1 )}

rather than a standard formula?

No standard formula covers reciprocals of products; partial fractions is the tool that manufactures the

f (k) - f (k + 1)

difference structure telescoping needs.

Why does summing the identity

(k + 1)^{3} - k^{3} = 3 k^{2} + 3 k + 1

help us find

\sum k^{2}

The left side telescopes to

(n + 1)^{3} - 1

; the right side is

3 \sum k^{2} + 3 \sum k + \sum 1

. Since we already know

\sum k = \frac{n ( n + 1 )}{2}

and

\sum 1 = n

, the only unknown is

\sum k^{2}

— so one collapse plus these known sums solves for it.

Why can we pull a constant

c

out of

\sum c a_{k}

but not out of

\sum a_{k}^{2}

in a similar way?

Factoring

c

uses the distributive law once per term (linear operation); squaring is nonlinear, so

\sum a_{k}^{2}

shares nothing with

(\sum a_{k})^{2}

or with pulling anything "out."

Why does Gauss's forwards-plus-backwards trick give

2 S

and not

S

We write the same sum twice (once reversed) and add them term by term, so the combined total is

S + S = 2 S

; each column then conveniently equals

n + 1

, giving

2 S = n (n + 1)

and

S = \frac{n ( n + 1 )}{2}

Why does telescoping make finding an infinite series limit easy?

Telescoping gives an exact closed form for the partial sum (e.g.

1 - \frac{1}{n + 1}

), so the limit is just what that expression tends to as

n \to \infty

— see Convergence of Series.

Why is telescoping called the finite version of the Method of Differences?

Both express a term as a difference

f (k) - f (k + 1)

so the sum "differences out"; the method of differences is the general framework, telescoping its concrete finite-sum instance.

Edge cases

What is

\sum_{k = 1}^{n} \frac{1}{k ( k + 1 )}

when

n = 1

It is the single term

\frac{1}{1 \cdot 2} = \frac{1}{2}

, and the formula

1 - \frac{1}{n + 1} = 1 - \frac{1}{2} = \frac{1}{2}

agrees — a good one-term sanity check.

What does

\sum_{k = m}^{n} a_{k}

mean if

m = n

It is a single term

a_{m}

(count

n - m + 1 = 1

). Sigma over one value is not "nothing" — it is exactly that one term.

By convention, what is an empty sum

\sum_{k = m}^{n} a_{k}

when

m > n

It equals

0

, the additive identity. There are no terms to add, and

0

keeps identities like

\sum_{1}^{n} = \sum_{1}^{p} + \sum_{p + 1}^{n}

consistent.

In a telescoping sum with a gap of

g

(denominator

k (k + g)

), how many survivors appear at each end?

Exactly

g

heads and

g

tails survive, because each cancellation is delayed by

g

terms — gap

1

leaves one, gap

2

leaves two, and so on.

What happens to

\sum_{k = 1}^{n} \frac{1}{k ( k + 1 )} = 1 - \frac{1}{n + 1}

n \to \infty

The tail

\frac{1}{n + 1} \to 0

, so the infinite series converges to

1

— telescoping hands us the exact limit with no extra work.

If a "telescoping" attempt leaves interior terms uncancelled, what has gone wrong?

The term was not truly written as

f (k) - f (k + 1)

(or

f (k) - f (k + g)

) — either the partial-fraction split has a wrong sign/constant, or the two pieces use functions that don't line up between consecutive rows.

Recall One-line self-test

Cover everything: can you state why products don't split, why the count needs $+ 1$ , and why only two ends survive a telescope? If any answer is a bare "because it just does," reread the refresher figures above before moving on.

Connections

Partial Fractions — the engine behind most "spot the error" sign traps.
Method of Differences — the general framework these edge cases live inside.
Convergence of Series — where the limiting-behaviour items point next.
Arithmetic Progressions — linearity traps recur there.