What is the interval of convergence of #sum_1^oo [(2^n)(x^n)]/sqrt(n) #?

We will use the result that:

#sum_(n=1)^oo 1/n^alpha" "# converges if and only if #alpha > 1#

Note that:

#(2^n)(x^n) = (2x)^n#

If #x >= 1/2# then:

#sum_(n=1)^oo ((2^n)(x^n)) / sqrt(n) >= sum_(n=1)^oo 1/sqrt(n)" "# which diverges.

If #0 <= x < 1/2# then #0 <= 2x < 1# and

#sum_(n=1)^oo (2x)^n / sqrt(n) <= sum_(n=1)^oo (2x)^n = 1/(1-2x)" "# converges.

So:

#sum_(n=1)^oo (2x)^n / sqrt(n)" "# is absolutely convergent for #-1/2 < x < 1/2#

Next, note that:

#1/sqrt(n) - 1/sqrt(n+1) = (sqrt(n+1) - sqrt(n))/(sqrt(n)sqrt(n+1))#

#color(white)(1/sqrt(n) - 1/sqrt(n+1)) = (sqrt(n+1) - sqrt(n))/(sqrt(n)sqrt(n+1)) * (sqrt(n+1) + sqrt(n))/(sqrt(n+1) + sqrt(n))#

#color(white)(1/sqrt(n) - 1/sqrt(n+1)) = 1/(sqrt(n)sqrt(n+1)(sqrt(n+1) + sqrt(n))#

#color(white)(1/sqrt(n) - 1/sqrt(n+1)) < 1/(2 n^(3/2))#

Hence if #x=-1/2# then:

#sum_(n=1)^oo (2x)^n / sqrt(n) = sum_(n=1)^oo (-1)^n/sqrt(n) < sum_(n=1)^oo 1/(2n^(3/2))" "# converges

If #x < -1/2# then #EE N# such that:

#abs(2x)^N > sqrt(N+1)/sqrt(N)#

and hence:

#sum_(n=1)^oo (2x)^n / sqrt(n)" "# diverges

(by comparison with a geometric series)