The radius of convergent of the power series sum_ (n=1)^oo x^n/(n*2^n)n=1xnn2n is?

The radius of convergent of the power series sum_ (n=1)^oo x^n/(n*2^n)n=1xnn2n is?
My answer is |x|<1.But I don't know whether that is correct.

1 Answer
Jun 25, 2018

22

Explanation:

Here's the standard approach:

Using the ratio test, the series converges absolutolely when the below limit is strictly less than 11, diverges if the limit is strictly greater than 11, and is inconclusive otherwise:

lim_(n->oo)|a_(n+1)/a_n|

where a_n=x^n/(n*2^n).

Compute the limit:

color(white)(=)lim_(n->oo)|x^(n+1)/((n+1)*2^(n+1))*(n*2^n)/x^n|

=lim_(n->oo)|(xn)/(2(n+1))|

=|x/2|

Thus, the series converges for |x|<2. Thus, the radius of converges is 2.

Additional Notes
But what about when the test is inconclusive, or |x|=2? For x=2, the series becomes the divergent harmonic series. For x=-2, the series becomes the conditionally convergent alternating harmonic series.

Thus, the interval of convergence is -2<=x< 2 (conditional convergence when x=-2 and absolute convergence otherwise).