# Ratio test question

• Feb 16th 2010, 12:21 AM
anderson
Ratio test question
hi everyone

im stuck with this question, need help to clarify..

Determine this series is absolutely convergent or diverges by using ratio test.

$\displaystyle \sum ^\infty_{n=1}$$\displaystyle \frac{(-1)^n e^\frac{1}{n}}{n^3} i manage to get this \displaystyle \frac{\frac{(-1)^{n+1} e^\frac{1}{n+1}}{(n+1)^3}}{\frac{(-1)^n e^\frac{1}{n}}{n^3}} \displaystyle \frac{-(n^3)e^\frac{1}{n+1}}{e^\frac{1}{n+1}(n+1)^3} how do i complete this working, out of ideas.need help. really appreciate all your help & guidance, thank you in advance for all your help & support. • Feb 16th 2010, 12:45 AM Prove It Quote: Originally Posted by anderson hi everyone im stuck with this question, need help to clarify.. Determine this series is absolutely convergent or diverges by using ratio test. \displaystyle \sum ^\infty_{n=1}$$\displaystyle \frac{(-1)^n e^\frac{1}{n}}{n^3}$

i manage to get this
$\displaystyle \frac{\frac{(-1)^{n+1} e^\frac{1}{n+1}}{(n+1)^3}}{\frac{(-1)^n e^\frac{1}{n}}{n^3}}$

$\displaystyle \frac{-(n^3)e^\frac{1}{n+1}}{e^\frac{1}{n+1}(n+1)^3}$

how do i complete this working, out of ideas.need help.

really appreciate all your help & guidance, thank you in advance for all your help & support.

You actually need to find $\displaystyle \lim_{n \to \infty}\left|\frac{a_{n + 1}}{a_n}\right|$

$\displaystyle = \lim_{n \to \infty}\left|\frac{\frac{(-1)^{n + 1}e^{\frac{1}{n + 1}}}{(n + 1)^3}}{\frac{(-1)^n e^\frac{1}{n}}{n^3}}\right|$

$\displaystyle = \lim_{n \to \infty}\frac{\frac{e^{\frac{1}{n + 1}}}{(n + 1)^3}}{\frac{e^{\frac{1}{n}}}{n^3}}$

$\displaystyle = \lim_{n \to \infty}\frac{n^3e^{\frac{1}{n + 1}}}{(n + 1)^3e^{\frac{1}{n}}}$

$\displaystyle = \lim_{n \to \infty}\frac{n^3e^{\frac{1}{n + 1} - \frac{1}{n}}}{(n + 1)^3}$

$\displaystyle = \lim_{n \to \infty}\frac{n^3e^{-\frac{1}{n(n + 1)}}}{(n + 1)^3}$

$\displaystyle = \lim_{n \to \infty}\frac{n^3}{(n + 1)^3e^{\frac{1}{n(n + 1)}}}$

Now yo might want to try using L'Hospital's Rule to simplify...
• Feb 16th 2010, 01:05 AM
anderson
thank you for guiding. got no idea how to use L Hospital rule to simplify this question, can some one help me.

thank you for all your help & guidance.
• Feb 16th 2010, 01:12 AM
HallsofIvy
There is no need to use L'Hopital's rule. Separate it into two parts:
$\displaystyle = \frac{n^3}{(n + 1)^3}e^{-\frac{1}{n(n + 1)}}$

Now, $\displaystyle \lim_{n\to\infty}\left(\frac{n}{n+1}\right)^3$ should be easy. As n goes to infinity, $\displaystyle \frac{1}{n(n+1)}$ goes to 0. I think you have a problem applying the ratio test here!
• Feb 16th 2010, 01:16 AM
anderson
$\displaystyle \lim_{n\to\infty}\left(\frac{n}{n+1}\right)^3$ = 1

is that correct? then how do i simplify the other limit , $\displaystyle e^{-\frac{1}{n(n + 1)}}$

need some guide & help, thank you in advance for all help & support.
• Feb 16th 2010, 01:21 AM
Prove It
Quote:

Originally Posted by HallsofIvy
There is no need to use L'Hopital's rule. Separate it into two parts:
$\displaystyle = \frac{n^3}{(n + 1)^3}e^{-\frac{1}{n(n + 1)}}$

Now, $\displaystyle \lim_{n\to\infty}\left(\frac{n}{n+1}\right)^3$ should be easy. As n goes to infinity, $\displaystyle \frac{1}{n(n+1)}$ goes to 0. I think you have a problem applying the ratio test here!

Wouldn't that mean that $\displaystyle e^{-\frac{1}{n(n + 1)}}$ tends to $\displaystyle e^0 = 1$?

So really, all you have to worry about is the $\displaystyle \left(\frac{n}{n + 1}\right)^3 = \left(1 - \frac{1}{n + 1}\right)^3$ which tends to $\displaystyle 1$.

Ohhh, so it all tends to $\displaystyle 1$ - so I agree with you that the ratio test fails...

• Feb 16th 2010, 01:27 AM
anderson
did you mean

$\displaystyle e^-\frac{1}{n^2+n}$
$\displaystyle e^-\frac{\frac{1}{n^2}}{\frac{n^2}{n^2}+\frac{n}{n^2} }$
$\displaystyle e^0$ = 1 ?

thank you for confirming, sorry to trouble everyone...

• Feb 16th 2010, 01:41 AM
Prove It
Quote:

Originally Posted by anderson
did you mean

$\displaystyle e^-\frac{1}{n^2+n}$
$\displaystyle e^-\frac{\frac{1}{n^2}}{\frac{n^2}{n^2}+\frac{n}{n^2} }$
$\displaystyle e^0$ = 1 ?

thank you for confirming, sorry to trouble everyone...

Surely if $\displaystyle -\frac{1}{n(n + 1)}$ tends to $\displaystyle 0$ then $\displaystyle e^{-\frac{1}{n(n + 1)}}$ tends to $\displaystyle e^0 = 1$...