For what values of x does the following series converge?

Question

∑_n=1∞xⁿ/√n

Answer 1

∑_n=1^∞xⁿ/√n

By limit ratiotest: lim n→∞ Ixⁿ⁺¹/√n+1·√n/xⁿI = IxI < 1 or -1 < x < 1.

If x = - 1 we get series ∑_n=1^∞(-1)ⁿ/√n converges by alternative series test.

If x = 1 we get series ∑_n=1^∞1/√n diverges as p-series for p = 1/2 <1.

So interval of convergence of given series: [- 1, 1)

Answer 2

∑[n=1,∞] xⁿ/√n

Use the ratio test:

lim_n->∞ |a_n+1/a_n|

= lim_n->∞ |[(xⁿ⁺¹)/√(n+1)]/[xⁿ/√n]|

= lim_n->∞ |[(xⁿ⁺¹)/√(n+1)]*[√n/xⁿ]|

= lim_n->∞ |x|*|√n/√(n+1)|

= |x|

Set the limit to be less than 1 to determine the interval of convergence:

|x| < 1

-1 < x < 1

Now, check if the function converges at the endpoints x=-1 and x=1

If x=-1:

∑[n=1,∞] (-1)ⁿ/√n converges by the Alternating Series Test since lim_n->∞ 1/√n = 0 and 1/√n > 1/√(n+1)

If x=1:

∑[n=1,∞] 1ⁿ/√n = ∑[n=1,∞] 1/√n = ∑[n=1,∞] 1/n^1/2, which diverges by the p-series test since p=1/2<1

Hence, the interval of convergence is -1 ≤ x < 1, or [-1,1) in interval notation.

Hope this helped!