Finding the Interval of Convergence for Power Series | Ratio Test Explained

[xtrubambinoxpr]

Find the interval of convergence for the power series [itex]\sum[/itex] [itex]\frac{x^n}{\sqrt{n}}[/itex]

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
[tex]\lim_{n \rightarrow \infty} \frac{a_n}{a_{n+1}}=\lim_{n \rightarrow \infty} \sqrt{\frac{n+1}{n}}=\cdots[/tex]

 

[xtrubambinoxpr]

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

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
[tex]\lim_{n \rightarrow \infty} \frac{a_n}{a_{n+1}}=\lim_{n \rightarrow \infty} \sqrt{\frac{n+1}{n}}=\cdots[/tex]

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