1. ## limits of series

using ratio test:

lim(n->inf) n^0.5 / (n+1)^0.5

=

lim(n->inf) 1 ?

2. Originally Posted by skett
using ratio test:

lim(n->inf) n^0.5 / (n+1)^0.5

=

lim(n->inf) 1 ?
ratio test is the (n+1)st term over the nth term ... might help to post the entire question.

3. sum[n=1] x^n/n^(1/2) ratio test

I can get it down to :

lim(n->inf) | x^(n+1) / (n+1)^(1/2) * n^(1/2) / x^n |

and simplify to :

lim(n->inf) | x n^(1/2) / (n+1)^(1/2) |

the book shows the answer of |x| and the only way I could think that would be possible is if the previous post were true.

4. Originally Posted by skett
sum[n=1] x^n/n^(1/2) ratio test

I can get it down to :

lim(n->inf) | x^(n+1) / (n+1)^(1/2) * n^(1/2) / x^n |

and simplify to :

lim(n->inf) | x n^(1/2) / (n+1)^(1/2) |

the book shows the answer of |x| and the only way I could think that would be possible is if the previous post were true.
understand that this problem results in an interval of convergence for the infinite series.

the limit is as $n \to \infty$ ... the absolute value of x can be factored out from the limit.

$\displaystyle |x| \lim_{n \to \infty} \left(\frac{n}{n+1}\right)^{\frac{1}{2}} < 1$

$|x| \cdot 1 < 1$

$|x| < 1$

$-1 < x < 1$

the final step is to check each endpoint for possible inclusion in the interval of convergence.

one more question ... is this problem from a "precalculus" course?

My guess is this: $\sum\limits_{n = 1}^\infty {\dfrac{{x^n }}{{\sqrt n }}}$

Then by either the root test or the ratio test the limit is $|x|$.
So we want $|x|<1$. So you must test for $x=-1\text{ or }x=1$.

6. Plato,
The page you are directing to produces 404.
I too found this link in the faq before posting, but as you can tell, it fails.

Skeeter,
I think I understand that you can factor out the abs(x) because x is any constant? And the result is convergent from -1<x<1 but is -1 not convergent as well [-1,1) ?

one more question ... is this problem from a "precalculus" course?
No, its from a "intermediate Calc" course, but I figured the question was basic in itself and would fit a precalc question.

7. x = -1 makes the series converge (alternating series w/ decreasing terms to 0), so the interval of convergence is -1 < x < 1

I believe this is a calculus topic ... I know of no precalculus courses which teach the concept of intervals of convergence for infinite series.

8. Hi can someone help me solve this convergence question attached? thanks!