Description

Devansh Jain
IIT Bombay
Questions to be Discussed
Sheet 1
2 (iv) – Sandwich Theorem for finding limits
3 (ii) – Checking convergence of a sequence
5 (iii) – Monotonic Bounded sequences are convergent
7 – Proof using definition
9 – Relation between product and convergence
11 – Conditions for exchanging product and limits (Discussed in lecture)
2. (iv) lim (n)1/n.
n→∞
Define hn := n1/n − 1.
Then, hn ≥ 0 ∀n ∈ N. (Why?)
Observe the following for n > 2 :
n = (1 + hn)n > 1 + nhn .
Thus, hn .
Using Sandwich Theorem, we get that lim hn = 0 which gives us that lim n1/n = 1.
n→∞ n→∞

3. (ii) To show: is not convergent.
n≥1
We will use the following two results: (a) Sum of convergent sequences is convergent.
(b) The sequence {(−1)n}n≥1 is not convergent. We now proceed as follows:
an := (−1)n .
(−1)n
It is easy to show that bn := is convergent. (Its absolute value will behave the n
same way as 1/n.)
Now, for the sake of contradiction, let us assume that (an) converges. Then, by (a),
(−1)n
we have it that cn := an + bn = must be convergent.
2
However, (cn) converging is equivalent to {(−1)n}n≥1 converging. (Why?) However, by (b), we know that the above is false. Thus, we have arrived at a contradiction.
an
5. (iii) a1 = 2, an+1 = 3 + ∀n ≥ 1.
2
Claim 1. an < 6 n ∈ N.
Proof. We shall prove this via induction. The base case n = 1 is immediate as 2 < 6. Assume that it holds for n = k. an 6
By principle of mathematical induction, we have proven the claim.
Claim 2. an < an+1 ∀n ∈ N.
a
Proof. an+1 − an = 3 − n = 6 − an > 0 =⇒ an+1 > an.
2 2
ak+1 = 3 + .
2
Thus, (an) is a monotonically increasing sequence that is bounded above. Therefore, it must converge. Use lim an+1 = lim an = L and then solve for L.
n→∞ n→∞
7. If lim an = L 6= 0, show that there exists n0 ∈ N such that n→∞
|L|
|an| ≥ for all n ≥ n0.
2
|L|
Let us choose . Why is this a valid choice of ?)
By hypothesis, there exists n0 ∈ N such that whenever n ≥ n0.

n0
9. (i) {anbn}n≥1 is convergent, if {an}n≥1 is convergent.
(ii) {anbn}n≥1 is convergent, if {an}n≥1 is convergent and {bn}n≥1 is bounded.
Both are false.
The sequences, an := 1 ∀n ∈ N and bn := (−1)n ∀n ∈ N act as a counterexample for both the statements.
11. (i) We shall show that the statement is false with the help of a counterexample.
Let a = −1, b = 1, c = 0. Define f and g as follows:
1/x ;x 6= 0 f (x) = x and g(x) = . 0 ;x = 0
It can be seen that lim f (x) = 0 but lim[f (x)g(x)] = lim 1 = 1. x→0 x→c x→0
(ii) We shall prove that the given statement is true.
We are given that g is bounded. Thus, ∃M ∈ R+ such that |g(x)| ≤ M ∀x ∈ (a, b).
Let > 0 be given. We want to show that there exists δ > 0 such that whenever 0 < |x − c| < δ.
Let . As lim f (x) = 0, there exists δ > 0 such that x→c
.
Thus, whenever 0 < |x − c| < δ, we have it that
(iii) We shall prove that the given statement is true.
Let > 0 be given. Let l := lim g(x).
x→c
Let .
By hypothesis, there exists δ1 > 0 such that 0
Also, there exists δ2 > 0 such that 0 .
Let δ = min{δ1, δ2}. Then, whenever 0 < |x − c| < δ, we have that:

Thus, we have it that 0

References