A solution to a possible QotD
We could use either the Ratio Test or the Root Test to analyze this
series. If an=(–1)nxn/sqrt(n)
then |an|=|x|n/sqrt(n). What about the
nth root? (I'll use the Root Test)? Well, the
|x|n will "give birth" to just |x|. The 1/sqrt(n) becomes
n–1/2 and as for the nth root:
(n–1/2)1/n=n–1/(2n)=(n1/n)–1/2
Before we recalled (first sequence lecture) that
n1/n→1 as n→∞. So combining this with the
algebra above, as n→∞ we know that
|an|1/n→|x|. So if x is between –1
and 1, the series converges absolutely and therefore it must
converge. If |x|>1 (so x<–1 or x>1) then the series
diverges.
Let's consider the situation at x=1. There we have
∑(–1)n/sqrt(n). In fact, this series converges
conditionally, since it converges (Alternating Series Test) but
the series of absolute values of the terms does not converge (p-series
with p=1/2<1). If x=–1, then the series becomes
∑n=1∞1/sqrt(n) (all the –'s
cancel!) and this series diverges. So:
The series converges absolutely in (–1,1) (interval of
convergence, so radius of convergence is 1), converges conditionally
if x=1, and diverges elsewhere.
Maintained by
greenfie@math.rutgers.edu and last modified 4/13/2009.