Sequences And Series Of Real Numbers -7 - Mathematics MCQ

The series Sigma [sin(1/n)] is a divergent series, because

lim(n-->inf.)[sin(1/n)]/(1/n) = 1, which is nonzero, which in turn is a consequence of lim(x-->0)[sin(x)]/x = 1. Hence the two series Sigma [sin(1/n)] and Sigma(1/n) have the same convergence behaviour by limit comparison test, for series of positive terms But we know that the harmonic series Sigma(1/n) diverges. Hence the given series Sigma[sin(1/n) also diverges.

Note that for all 1 </= n, 0 < (1/n) </= 1 < π, and so sin(1/n) is positive i.e. the given series is of positive terms.

Correct Answer :- A

Explanation : The series for Rabee’s test is :

Converge when there exists a c>1 such that l is greater than and equal to c for all n>N.

Diverge when l is less than and equal to 1 for all n>N.

Otherwise, the test is inconclusive.

Here power series is


 

The given series is : x - x²/2 + x³/3 - x⁴/4.........(A)

The series formed by the coefficients is : 

1 - 1/2 + 1/3 - 1/4.........(B)By I the series is absolutely convergent when x lies between −1 and +1  

By II the series is divergent when x is less than −1 or greater than +1

By III there is no test when x = ±1.  

Thus the given series is said to have [−1, 1] as the interval of convergence.