Determine the radius and interval of convergence

∞
Σ ( (-1)^k / √ k ) * (3x-1)^k
k=1

Let Ak = ( (-1)^k / √ k ) * (3x-1)^k
|Ak| = (3x-1)^k / √ k
If |Ak| converges, An converges absolutely.

Use nth, or kth in this case, root test for |Ak|.
If kth root of |Ak| < 1 as k tends to inf.,
|Ak| converges.

Kth root of |Ak|
= |3x - 1| / k^(1/2k)

lim (k->inf) |3x - 1| / k^(1/2k)
= |3x - 1| ----------- lim (k->inf) k^(1/2k) = 1, can be obtained by taking ln.

Let |3x-1| < 1
-1 < 3x - 1 < 1
0 < x < 2/3

Now you have to check end points,
Namely, when 3x - 1 = 1 or -1

When 3x-1 = 1, An becomes an converging alternating series.
When 3x-1 = -1, An is a diverging p series.

Thus, the interval of convergence is 0 < x < = 2/3
The radius = (2/3 - 0) / 2 = 1/3