Question

examine the behaviour of infinite series of positive terms
(1/2)²+ (1 • 3 / 2 • 4)² + (1 • 3 •5 / 2 • 4 • 6)²​

Answer 1

Answer: Let (an) be a sequence of real numbers. Then an expression of the form a1 +a2 +a3 +......

denoted by P∞

n=1 an, is called a series.

Examples : 1. 1 + 1

2 +

1

3 + .... or P∞

n=1

1

n

2. 1 + 1

4 +

1

9 + .... or P∞

n=1

1

n2

Partial sums : Sn = a1 + a2 + a3 + ...... + an is called the nth partial sum of the series P∞

n=1 an,

Convergence or Divergence of P∞

n=1 an

If Sn → S for some S then we say that the series P∞

n=1 an converges to S. If (Sn) does not

converge then we say that the series P∞

n=1 an diverges.

Examples :

1. P∞

n=1 log(

n+1

n

) diverges because Sn = log(n + 1).

2. P∞

n=1

1

n(n+1) converges because Sn = 1 −

1

n+1 → 1.

3. If 0 < x < 1, then the geometric series P∞

n=0 x

n

converges to 1

1−x

because Sn =

1−x

n+1

1−x

.

Necessary condition for convergence

Theorem 1 : If P∞

n=1 an converges then an → 0.

Proof : Sn+1 − Sn = an+1 → S − S = 0. ¤

The condition given in the above result is necessary but not sufficient i.e., it is possible that

an → 0 and P∞

n=1 an diverges.

Examples :

1. If | x | ≥ 1, then P∞

n=1 x

n diverges because an 9 0.

2. P∞

n=1 sinn diverges because an 9 0.

3. P∞

n=1 log(

n+1

n

) diverges, however, log(

n+1

n

) → 0.

Necessary and sufficient condition for convergence

Theorem 2: Suppose an ≥ 0 ∀ n. Then P∞

n=1 an converges if and only if (Sn) is bounded above.

Proof : Note that under the hypothesis, (Sn) is an increasing sequence. ¤

Example : The Harmonic series P∞

n=1

1

n

diverges because

S2

k ≥ 1 +

1

2

+ 2 ·

1

4

+ 4 ·

1

8

+ ... + 2k−1

·

1

2

k

= 1 +

k

2

for all k.

Theorem 3: If P∞

n=1 | an | converges then P∞

n=1 an converges.

Proof : Since P∞

n=1 | an | converges the sequence of partial sums of P∞

n=1 | an | satisfies the Cauchy

criterion. Therefore, the sequence of partial sums of P∞

n=1 an satisfies the Cauchy criterionLet (an) be a sequence of real numbers. Then an expression of the form a1 +a2 +a3 +......

denoted by P∞

n=1 an, is called a series.

Examples : 1. 1 + 1

2 +

1

3 + .... or P∞

n=1

1

n

2. 1 + 1

4 +

1

9 + .... or P∞

n=1

1

n2

Partial sums : Sn = a1 + a2 + a3 + ...... + an is called the nth partial sum of the series P∞

n=1 an,

Convergence or Divergence of P∞

n=1 an

If Sn → S for some S then we say that the series P∞

n=1 an converges to S. If (Sn) does not

converge then we say that the series P∞

n=1 an diverges.

Examples :

1. P∞

n=1 log(

n+1

n

) diverges because Sn = log(n + 1).

2. P∞

n=1

1

n(n+1) converges because Sn = 1 −

1

n+1 → 1.

3. If 0 < x < 1, then the geometric series P∞

n=0 x

n

converges to 1

1−x

because Sn =

1−x

n+1

1−x

.

Necessary condition for convergence

Theorem 1 : If P∞

n=1 an converges then an → 0.

Proof : Sn+1 − Sn = an+1 → S − S = 0. ¤

The condition given in the above result is necessary but not sufficient i.e., it is possible that

an → 0 and P∞

n=1 an diverges.

Examples :

1. If | x | ≥ 1, then P∞

n=1 x

n diverges because an 9 0.

2. P∞

n=1 sinn diverges because an 9 0.

3. P∞

n=1 log(

n+1

n

) diverges, however, log(

n+1

n

) → 0.

Necessary and sufficient condition for convergence

Theorem 2: Suppose an ≥ 0 ∀ n. Then P∞

n=1 an converges if and only if (Sn) is bounded above.

Proof : Note that under the hypothesis, (Sn) is an increasing sequence. ¤

Example : The Harmonic series P∞

n=1

1

n

diverges because

S2

k ≥ 1 +

1

2

+ 2 ·

1

4

+ 4 ·

1

8

+ ... + 2k−1

·

1

2

k

= 1 +

k

2

for all k.

Theorem 3: If P∞

n=1 | an | converges then P∞

n=1 an converges.

Proof : Since P∞

n=1 | an | converges the sequence of partial sums of P∞

n=1 | an | satisfies the Cauchy

criterion. Therefore, the sequence of partial sums of P∞

n=1 an satisfies the Cauchy criterion

Step-by-step explanation:

Answer 2

Answer:

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