Does this sequence converge or diverge?

**linalg123** · September 10th 2012, 05:27 AM

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

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

**emakarov** · September 10th 2012, 05:40 AM

I recommend calculating the first several elements and making a conjecture.

**Prove It** · September 10th 2012, 05:46 AM

It definitely does not converge. $\displaystyle \begin{align*} \frac{n}{n + 1} \to 1 \end{align*}$ as $\displaystyle \begin{align*} n \to \infty \end{align*}$ , so eventually the terms will oscillate between -1 and 1.

**amillionwinters** · September 10th 2012, 05:56 AM

Does this help you?

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

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

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

$\rightarrow \lim_{n\to\infty}(-1)^n(1)$

**linalg123** · September 10th 2012, 07:04 AM

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.

**Plato** · September 10th 2012, 07:49 AM

Originally Posted by linalg123

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.

If $f_n=\frac{(-1)^nn}{n+1}$ and $n>1$ then it can be shown that $f_n>\tfrac{1}{2}$ if $n$ even and $f_n<-\tfrac{1}{2}$ if $n$ is odd.

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

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