Sequences And Series Of Real Numbers -4
20 Questions MCQ Test Topic-wise Tests & Solved Examples for IIT JAM Mathematics | Sequences And Series Of Real Numbers -4
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thenCorrect Answer :- D
Explanation : For |a|<1
b = 1/a, |b|>1
lim (n→∞) an = lim (n→∞) (1/b)n
= lim(n→∞) 1/bn
= 1/(±∞)
= 0
Which amongst the following statements is true?
Option c will be correct answer of this . every bounded sequence is not convergent for example (-1)^n is bounded by -1 and 1 but it is not convergent .
A Cauchy sequence is convergent, if it is a
If radius of convergence of series is 1, then radius of convergence of series
is…
By the differentiation’s theorem, we have
...(i)
Differentiating term by term, we obtain
...(ii)
So, both the series have same radius of convergence.
If {xn} and {yn} are two convergent sequence such that xn < yn, 
n ∈ N, then
The correct option is Option A.
Let lim xn = x and lim yn = y and zn = yn - xn
n->∞ n->∞
Then (zn) is a convergent sequence such that zn > 0 ∇ n ∊N
and lim zn = y - x.
n->∞
Now (zn) is a convergent sequence of real numbers and zn > 0 ∇ n ∊N
So, lim zn ≥ 0
n->∞
So, y - x ≥ 0
=> y ≥ x
=> lim yn ≥ lim xn
n->∞ n->∞
Hence, proved.
Let a = min{x2 + 2x + 3, x ∈ R} and 
then the value of 
Correct Answer :- b
Explanation : Suppose otherwise, that there exists a number L implies R and a positive integer N such that
| f(n) - L | < e {for all } e > 0 {for all } n > N.
Since N is a positive integer, we know 4N > N and 4N+2 > N.
But,
f(4N) = cos (2N pi) = 1 and f(4N+2) = cos((2N+1)pi) = -1
Taking e = 1/2
= |1 - L| < 1/2 and |-1-L| < 1/2
=> |1+L| < 1/2
But, these imply
|1-L| + |1+L| < 1.
By the triangle inequality we then have
|1 - L + 1 + L | < 1
=> 2 < 1
a contradiction. Hence, there is no such limit L.
Therefore, the sequence converges to zero.
If for any e > 0, there exists a positive integer m such that I an - am | < e whenever n ≥ m . The sequence an > is called
A convergent sequence is a Cauchy sequence, if it is a
ANSWER :- b
Solution :- If {an}∞n=1 is a cauchy sequence of real numbers and if there is a sub-sequence of this sequence, {anj}∞j=1 which converges to a real number L, then I need to show that the sequence {an}∞n=1 converges to the real number L.
For the given Arithmetic progression find the position of first negative term? 50, 47, 44, 41,............
Let nth term=0, the next term would be first negative term.
0 = 50 + (n - 1) – 3, n = 17.66.. therefore at n = 18 the first negative term would occur.
In the given AP series the term at position 11 would be? 5, 8, 11, 14, 17, 20.........50.
nth term = a + (n – 1)d, nth term = 5 + (11 - 1)3 = 35.
The value of K for which the set of equations x + ky + 3z = 0, 3x + ky - 2z = 0, 2x + 3y - 4z = 0, has a non-trivial solution over the set of rationales is
For non-trivial soln det of Coefficient, matrix should be 0.
|1 K 3|
|3 K 2| = 0
|2 3 4|
4K - 4K + 27 - 6K + 6 + 12K = 0
2K + 33 = 0
33 K=2
K=33/2
are convergent, thenA sequence contains a convergent subsequence, if it is
be a sequence converges to 0 and 
be a sequence that is bounded, then 
is a sequence that
Which of the following sequences of functions is uniformly convergent on (0, 1)?
The sequence {xn}, where xn = nl/n, converge to
Let xn = 22n 
for all n ∈ N. Then the sequence {xn}
xn = 22n for all n ∈ N
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