The limit of a function makes sense if x is defined in the neighbourhood of a. The limit of a sequence makes only when variable approaches ∞ . A sequence a 1 , a 2 , a 3 ,.... of real numbers is said have a limit I, if 
If I is finite the sequence is said to be convergent. If I = ∞ , the sequence is said to be divergent. If a n does not approach a definite number, then the sequence {a n } is oscillatory.If above results are well known.
(i) Which of the following sequences does not converge to zero?
1 + e 2
e 
1 1
(ii) Let a 1 = 1, a n = n(a n–1 + 1) for n = 2, 3, ..., where P n = 
, then 
must be- ∞ (iii) Let {a n } be sequence of real numbers defined as a 1 = I, 
= 
for n ≥ 1, then 
must be- 0
Text Solution
Verified by ExpertsA
Ans.
(i)
Sol. The sequence given in choice converges to zero since after a certain stage 
will become very small as n → ∞ .
Let us prove :
Choose a natural number k > 2x, then for
n > k 
= 
= 
→ 0 as n → ∞ .
⇒ 
will be less than arbitrarily chosen small positive numbers after some stage.
It is not true since n approaches ∞ much faster than log n.
It is wrong since x n → 0 as n → ∞ (Recall infinite GP)
It is correct not only due to the fact that other choices have been eliminated but by the fact that n 1/n will always be greater than n whatever be the value of x. Thus is the correct choice.
(ii)
Sol.a 1 = 1, a 2 = 2(a n–1 + 1) = 4, a 3 = 3(4 + 1) = 15
P 1 = 2 = 1 + 
, P 2 = 
= 
= 
= 1 + 
P 3 = 
= 
= 
and so on. ⇒ 
(iii)
Sol. a 2 2 = 
= 
⇒ a 2 = 
> 1
⇒ a 2 > a 1
a n 2 = 
⇒ 
= 
....(i)
Since a 2 > a 1 , from Eq.(1), a 3 > a 2 and so on.
Thus 
for all n.
⇒ 
⇒ 
< 2 for all n
⇒ 1 < a n < 2 for all n
Since {a n } is increasing.
∴ 
= 
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems