Interval of convergence for ∑x^n/√n using the ratio test

[xtrubambinoxpr]

Find the interval of convergence for the power series $\sum$ $\frac{x^n}{\sqrt{n}}$

using the ratio test I get that the absolute value of x * the lim of square root of n over square root of n+1 = 0. so that being said i believe the interval of convergence is (-∞,∞) by the ratio test

 

[vanhees71]

Think about your limit again! The convergence radius is
$$\lim_{n \rightarrow \infty} \frac{a_n}{a_{n+1}}=\lim_{n \rightarrow \infty} \sqrt{\frac{n+1}{n}}=\cdots$$

 

[xtrubambinoxpr]

Think about your limit again! The convergence radius is
$$\lim_{n \rightarrow \infty} \frac{a_n}{a_{n+1}}=\lim_{n \rightarrow \infty} \sqrt{\frac{n+1}{n}}=\cdots$$

Why? using the ratio test you get X^n+1 / sqrt(n+1) * the reciprocal of the original expression. So the x^n cancel out. leaving it in the form with n / n+1 square root

 

[xtrubambinoxpr]

Think about your limit again! The convergence radius is
$$\lim_{n \rightarrow \infty} \frac{a_n}{a_{n+1}}=\lim_{n \rightarrow \infty} \sqrt{\frac{n+1}{n}}=\cdots$$

also i notice you have n+1 in the denominator. Isnt it the numerator in the original formula? that's how it is in my book anyways