The series 1 · x + 1 · 2x2 + 1 · 2 3x2 + ... + + ... + n! · xn + ... is
Correct Answer :- a
Explanation : an = n!
then |an+1/an| = (n+1)/n!
L = lim(n-->∞) |an+1/an|
The radius of convergence R = 0 Hence, the series converges only at x = 0.
Option A) is correct, it is divergent.
The series 2 + 4 + 6 + 8 + ...is
The scries x is convergent, if
If ∑μn be the given series, then we have
∴ From ratio test, the given series ∑μn is convergent or divergent according as
If a1, a2, a3, ... be a sequence of non-zero integers satisfying
II. an divides an + 1
III. every integer is a divisor of some an, then
is a contradiction chair, and so, A is an irrational number.
Let ∑μn be a series of positive terms. Given that ∑μn is convergent and also exists, then the said limit is
Then, from Eq. (i) we have
Here, given series is
∴ If ∑μn be the given series, then we have
Therefore, from ratio test, we conclude that the given series ∑μn is convergent or divergent according as
i. e., according as x < 1 or > 1 ...(i)
If n = 1, this test fails and the given series reduces to
∑μn whose nth term and can be proved to be divergent.
Hence, we conclude that the given series is convergent, if x2 ≤ 1 and divergent, if x2 ≥ 1.
Match list I with list II and select the correct answer
is also convergent series.
∴ un is also convergent series.
Hence, by d’Alembert test
implies un is divergent series.
By Cauchy condensation test, ∑μn is convergent series.
By Leibnitz’s test, the series is convergent.
Also, is divergent series.
So, the given series is converges conditionally.
The series whose nth term is
Here it is given that
Now, taking auxiliary series
Which is finite and non-zero. Since, vn is divergent.
Hence, tn is also divergent.
The series convergent, if
Here given series is is convergent
∴ The given series will be convergent, if
Hence, vn is convergent series, so un is also convergent.
∴ The given series is convergent, if | x | ≤ 1.
Let the given series be denoted by ∑μn, then
Compare this series ∑vn with the auxiliary series,
which is finite quantity.
Hence, ∑un and ∑wn are either both convergent or both divergent. But ∑wn
Hence, the series ∑vn is divergent.
Also, in the series ∑un, we find that its term are alternately positive and negative, its terms are continuously decreasing and
Thus, all condition of Leibnitz’s test are satisfied and as such ∑un is convergent. H ence, the given series ∑un is conditionally convergent.
which is finite and non-zero.
Since, ∑vn is divergent, therefore, ∑un is also divergent.
Hence, by Cauchy’s root test, the series convergent.
Here, it is given that
Hence, ∑vn is divergent and ∑un is also divergent and
Hence, by d’Alembert’s ratio test, it is convergent.
The series convergent, if
The nth term of the given series is
Hence, ∑un is convergent.