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

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

1 Answer
NickTheTurtle
Jun 25, 2018

#2#

Explanation:

Here's the standard approach:

Using the ratio test, the series converges absolutolely when the below limit is strictly less than #1#, diverges if the limit is strictly greater than #1#, 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).

Related questions
Impact of this question
1510 views around the world
You can reuse this answer
Creative Commons License