Question

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

Answer 1

`[-1/2, 1/2)`

Explanation:

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)