# Thread:

1. ## Ratio and Root Tests

Good morning All,

I am working on ratio and root tests for infinite series. I would like to post what I have done, and see if anyone agrees with it. Please see below:

$<br /> \displaystyle\sum_{n=2}^{\infty}(-1)^n\frac{123^n}{n!}<br />$

By the Ratio Test...

$<br /> \displaystyle\lim_{n\rightarrow\infty}\mid\frac{(-1)^{(n+1)}123^{(n+1)}}{(n+1)!}\frac{n!}{(-1)^{n}123^{n}}\mid<br />$

And...

$<br /> \displaystyle\lim_{n\rightarrow\infty}\frac{123n!} {(n+1)!} = \lim_{n\rightarrow\infty}\frac{123n!}{(n+1)n!} = \lim_{n\rightarrow\infty}\frac{123}{n+1} = 0<br />$

Therefore...

The orignal series $\displaystyle\sum_{n=2}^{\infty}(-)^n\frac{123^n}{n!}$ is Absolutely Convergent.

Are we good here?

My second problem is:

$\displaystyle\sum_{n=1}^{\infty}(-1)^n(\sqrt{n+1}-\sqrt{n})$

I'm not really sure where to start here. Would I apply the root test to this problem?

2. The first problem looks good to me, and you have the right idea for the second. What do you get when you try to take the $n$-th root?

3. Thanks!

When I try to take the nth root i get the following:

$\displaystyle\sum_{n=1}^{\infty}(-1)^n(\sqrt{n+1}-\sqrt{n})$

$\displaystyle\lim_{n\rightarrow\infty}|[(-1)^n(\sqrt{n+1}-\sqrt{n})]^{\frac{1}{n}}|$

$=\displaystyle\lim_{n\rightarrow\infty}|(-1)^{\frac{n}{n}}((n+1)^{\frac{1}{2n}}-(n)^{\frac{1}{2n}})|$

$=\displaystyle|(-1)^1((n+1)^0-(n)^0)|$

$=\displaystyle|(-1)((1)-(1))|$

$=\displaystyle|-1+1|$

$=\displaystyle|0|$

$=\displaystyle0$

That can't be right??? I have three problems; one is supposed to be divergent, one conditionally convergent, and one absolutely convergent. Since my first problem was absolutely convergent (and I was able to follow the entire process for the first problem) I would have thought that the second problem would be either conditonally convergent or divergent.

...so, where did I go wrong?

4. Have you studied the alternating series test? If so that is what is to be used.

5. I have reviewed the alternating series test.

I might be making this harder on myself than it needs to be, but since the section deals with ratio and root tests, I would like to use one of those methods on this problem.

Is this reasonable?

6. Originally Posted by MechEng
I have reviewed the alternating series test.

I might be making this harder on myself than it needs to be, but since the section deals with ratio and root tests, I would like to use one of those methods on this problem.
Is this reasonable?
Did you notice that $\displaystyle x_n=(-1)^n\left(\sqrt{n+1}-\sqrt{n}\right)=\frac{(-1)^n}{\sqrt{n+1}+\sqrt{n}\right }?$

7. Originally Posted by Plato
Did you notice that $\displaystyle x_n=(-1)^n\left(\sqrt{n+1}-\sqrt{n}\right)=\frac{(-1)^n}{\sqrt{n+1}+\sqrt{n}\right }?$
Shoot... I did not notice that. Let me try this again...

8. When I try to take the nth root i get the following:

$\displaystyle\sum_{n=1}^{\infty}(-1)^n(\sqrt{n+1}-\sqrt{n}) = \displaystyle\sum_{n=1}^{\infty}\frac{(-1)^n}{(\sqrt{n+1}+\sqrt{n})}$

$=\displaystyle\lim_{n\rightarrow\infty}|[\frac{(-1)^n}{(\sqrt{n+1}+\sqrt{n})}]^{\frac{1}{n}}|$

$=\displaystyle\lim_{n\rightarrow\infty}|\frac{(-1)^\frac{n}{n}}{(n+1)^\frac{1}{2n}+(n)^\frac{1}{2n }}}|$

$=\displaystyle|\frac{(-1)^1}{(n+1)^0+(n)^0}|$

$=\displaystyle|\frac{-1}{1+1}| = \frac{1}{2}$

Am I losing it again somewhere?

Or, are you implying that I use the Ratio Test?

Thanks for the help.

9. Use the alternating series test.

10. Ok, I am waiting to hear back with respect to whether or not using the alternating series test is acceptable.

In the interim, I have worked out the last of three problems and concluded that:

$\displaystyle\sum_{n=1}^{\infty}(-1)^n\frac{(2n)!}{3^n(n!)}$

By Ratio Test...

$\displaystyle\lim_{n\rightarrow\infty}|\frac{(-1)^{(n+1)}(2(n+1))!}{3^{(n+1)}((n+1)!)}\frac{3^n(n !)}{(-1)^n(2n)!}|$

$=\displaystyle\lim_{n\rightarrow\infty}\frac{4n^2+ 6n+2}{3n+3} = \infty$

Therefore...

By the Ratio test, the series Diverges

So, this leaves me with my remaining series that is conditionally convergent. And, this means that if I am able to use the Ratio or Root Test, I should wind up with a limit equal to one... right?

11. Plato,

I am noticing a trend with the lesson plans that I am following; the questions seem to be quite difficult using the methods covered in that section, but following section will introduce a method that makes the problem quite simple to do. I assume this one is the same.

I am not a huge fan of the way the text book is organized, but i am nearly done with it so it's not worth complaining now.

I appreciate your patience. I used to tutor physics in college, so I know it can be trying when someone keeps mising the fundamental ideas. Thank you for continuing to work with me.

12. All tests are now available to use on this problem, so I will heed you advice, Plato, and try the alternating series test...

13. You should find that it in conditional but not absolute.
To do the later, use the limit comparison test using $\dfrac{1}{\sqrt{n}}$

14. Easy question I hope, but...

Since one of the requirements for the limit comparison test is that the series are both positive, does that mean that my original series becomes...

$\displaystyle\sum_{n=1}^{\infty}\frac{1}{\sqrt{n+1 }+\sqrt{n}}$

15. Please correct me if I am out of whack here...

Let $\displaystyle\sum a_n = \frac{1}{\sqrt{n+1}+\sqrt{n}}$

So...

$\displaystyle\lim_{n\rightarrow\infty}\frac{1}{\sq rt{n+1}+\sqrt{n}}=0$

By Alternating Series Test... Convergent

Let $\displaystyle\sum b_n = \frac{1}{\sqrt{n}}$ A Divergent P-series

So...

$\displaystyle\lim_{n\rightarrow\infty}\frac{1}{\sq rt{n+1}+\sqrt{n}}\frac{\sqrt{n}}{1}$

$=\displaystyle\lim_{n\rightarrow\infty}\frac{1}{\s qrt{n+1}}=0$

By Limit Comparison Test... Divergent

This doesn't make sense, does it?

Would it make ore sense if the order of the tests was reversed?
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 31, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9530503153800964, "perplexity_flag": "middle"}