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Correct Answer : D
Explanation : For a<1
b = 1/a, b>1
lim (n→∞) a_{n} = lim (n→∞) (1/b)^{n}
= lim(n→∞) 1/b^{n}
= 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 {x_{n}} and {y_{n}} are two convergent sequence such that x_{n} < y_{n}, 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{x^{2} + 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 1L < 1/2
=> 1+L < 1/2
But, these imply
1L + 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 a_{n}  a_{m}  < e whenever n ≥ m . The sequence a_{n} > 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 subsequence 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 n^{th} 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.
n^{th} term = a + (n – 1)d, n^{th} 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 nontrivial solution over the set of rationales is
For nontrivial 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
A 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 {x_{n}}, where x_{n} = n^{l/n}, converge to
Let x_{n }= 2^{2n} for all n ∈ N. Then the sequence {x_{n}}
x_{n} = 2^{2n} for all n ∈ N
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