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CHAPTER 2

INFINITE SUMS (SERIES)

Lecture Notes

We extend the notion of a finite sum Σnk=1 ak to an INFINITE SUM which we write as

Σ∞ n=1 an as follows.

DEFINITION 1

For a given sequence {an}n∈N −{ 0 }, i.e the sequence a 1 , a 2 , a 3 , ....an, .....

we form a following (infinite) sequence

S 1 = a 1 , S 2 = a 1 +a 2 , ...., Sn = Σnk=1 ak, .......

We use it to define the infinite sum as fol- lows.

DEFINITION 2

If the limit limn→∞ Sn exists and is finite, i.e.

nlim→∞ Sn^ =^ S,

then we say that the infinite sum Σ∞ n=1 an CONVERGES to S and

we write it as

Σ∞ n=1 an = S,

otherwise the infinite sum DIVERGES.

In a case that

n^ lim→∞ Sn exists and is infinite, i.e.

nlim→∞ Sn^ =^ ∞,

we say that the infinite sum

Σ∞ n=1 an DIVERGES to ∞ and

we write it as

Σ∞ n=1 an = ∞.

In a case that limn→∞ Sn does not exist we say that the infinite sum Σ∞ n=1 an DIVERGES.

EXAMPLE 2 The series Σ∞ n=1 1 DIVERGES to ∞ as Sn = Σnk=11 = n and

nlim→∞ Sn^ =^ nlim→∞ n^ =^ ∞.

EXAMPLE 3 The infinite sum Σ∞ n=1 (−1)n DIVERGES.

EXAMPLE 4 The infinite sum

Σ∞ n=

(k + 1)(k + 2)

CONVERGES and

Σ∞ n=

(k + 1)(k + 2)

Proof: first we evaluate Sn = Σnk=1 (^) (k+1)(^1 k+2) as follows. Sn = Σnk=1^1 (k + 1)(k + 2)

= Σnk=1k−^2 =

x + 1

|n 0 +1 = −

n + 2

• 1 and

nlim→∞ Sn^ =^ nlim→∞ −^

n + 2

General Properties of Infinite Sums

THEOREM 1

If Σ∞ n=1an converges, then

nlim→∞ an^ = 0.

Proof: observe that an = Sn − Sn− 1 and hence

nlim→∞ an^ =^ nlim→∞ Sn^ T he^ −^ nlim→∞ Sn−^1 = 0, as limn→∞ Sn = limn→∞ Sn− 1.

REMARK The reverse statement to the the- orem 1 If (^) nlim→∞ an = 0. then Σ∞ n=1an converges

is not always true. There are infinite sums with the term converging to zero that are not convergent.

EXAMPLE 5 The infinite HARMONIC sum

H = Σ∞ n=

n DIVERGES to ∞, i.e.

Σ∞ n=

n

but limn→∞ (^1) n = 0.

THEOREM 2 If the infinite sums

Σ∞ n=1an, Σ∞ n=1bn CONVERGE, then the following properties hold.

Σ∞ n=1(an + bn) = Σ∞ n=1an + Σ∞ n=1bn,

Σ∞ n=1can = cΣ∞ n=1an, c ∈ R.

Alternating Infinite Sums

DEFINITION 5 An infinite sum

Σ∞ n=1(−1)n+1an, for an ≥ 0 is called ALTERNATING infinite sum (al- ternating series).

EXAMPLE 6 Consider

Σ∞ n=1(−1)n+1^ = 1 − 1 + 1 − 1 + .... If we group the terms in pairs, we get

(1 − 1) + (1 − 1) + .... = 0 but if we start the pairing one step later, we get

1 −(1−1)−(1−1)−..... = 1− 0 − 0 − 0 −... = 1.

THEOREM 3 The alternating infinite sum

Σ∞ n=1(−1)n+1an, (an ≥ 0) such that a 1 ≥ a 2 ≥ a 3 ≥ .... and (^) nlim→∞ an = 0 always CONVERGES.

Moreover, its partial sums

Sn = Σnk=1(−1)n+1an fulfil the condition

S 2 n ≤ Σ∞ n=1(−1)n+1an ≤ S 2 n+1.

Proof: observe that the sequence of S 2 n is in- creasing as

S 2 n+2 = S 2 n + (a 2 n+1 − a 2 n+

and a 2 n+1 − a 2 n+2 ≥ 0 ,

i.e. S 2 n+2 ≥ S 2 n.

The sequence of S 2 n is also bounded as

S 2 n = a 1 −((a 2 −a 3 )+(a 4 −a 5 )+...a 2 n) ≤ a 1.

We know that any bounded and increasing sequence is is convergent, so we proved that S 2 n converges.

We prove in a similar way that the sequence {S 2 n+1} is decreasing.

Hence we get

S 2 n ≤ (^) nlim→∞ S 2 n = g = Σ∞ n=1(−1)n+1an and S 2 n+1 ≥ (^) nlim→∞ S 2 n+1 = g

and Σ∞ n=1(−1)n+1an = g,

i.e S 2 n ≤ Σ∞ n=1(−1)n+1an ≤ S 2 n+1.

EXAMPLE 7

Consider the ANHARMONIC series AH = Σ∞ n=1 (−1)n+1^1 n

Observe that an = (^1) n, and

1 n

n + 1 i.e. an ≥ an+1 for all n.

This proves that the assumptions of the the- orem 3 are fulfilled for AH and hence AH CONVERGES.

In fact, it is proved (by analytical methods) that AH = Σ∞ n=1(−1)n+^

n

= ln 2.