# 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.

$\sum ^\infty_{n=1}$ $\frac{(-1)^n e^\frac{1}{n}}{n^3}$

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

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

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.

$\sum ^\infty_{n=1}$ $\frac{(-1)^n e^\frac{1}{n}}{n^3}$

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

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

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 $\lim_{n \to \infty}\left|\frac{a_{n + 1}}{a_n}\right|$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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