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

If it converges find the limit and if it diverges justify why.

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- Sep 10th 2012, 05:27 AMlinalg123Does this sequence converge or diverge?
$\displaystyle f= (-1)^n \frac{n}{n+1}$

If it converges find the limit and if it diverges justify why. - Sep 10th 2012, 05:40 AMemakarovRe: Does this sequence converge or diverge?
I recommend calculating the first several elements and making a conjecture.

- Sep 10th 2012, 05:46 AMProve ItRe: Does this sequence converge or diverge?
It definitely does not converge. $\displaystyle \displaystyle \begin{align*} \frac{n}{n + 1} \to 1 \end{align*}$ as $\displaystyle \displaystyle \begin{align*} n \to \infty \end{align*}$, so eventually the terms will oscillate between -1 and 1.

- Sep 10th 2012, 05:56 AMamillionwintersRe: Does this sequence converge or diverge?
Does this help you?

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

$\displaystyle \rightarrow \lim_{n\to\infty}(-1)^n\frac{n/n}{n/n+1/n}$

$\displaystyle \rightarrow \lim_{n\to\infty}(-1)^n\frac{1}{1+0}$

$\displaystyle \rightarrow \lim_{n\to\infty}(-1)^n(1)$ - Sep 10th 2012, 07:04 AMlinalg123Re: Does this sequence converge or diverge?
Thanks for the replies. I understand the sequence definitely diverges.

Please note my brother asked me this question and it is not part of my coursework(and i have never done anything like this before), which is why i didn't make an attempt at a solution, however i am still interested in the answer.

His question asked to find an appropriate epsilon and n to justify why the series diverges. - Sep 10th 2012, 07:49 AMPlatoRe: Does this sequence converge or diverge?
If $\displaystyle f_n=\frac{(-1)^nn}{n+1}$ and $\displaystyle n>1$ then it can be shown that $\displaystyle f_n>\tfrac{1}{2}$ if $\displaystyle n$ even and $\displaystyle f_n<-\tfrac{1}{2}$ if $\displaystyle n$ is odd.

That means $\displaystyle |f_{n+1}-f_n|>1.$