# Prove wether or not the following series diverges or converges: $\sum_{n=0}^\infty {(-1)^nn\over n+1}$

Prove wether or not the following series diverges or converges:

$\sum_{n=0}^\infty {(-1)^nn\over n+1}$

I am just not sure, I know if I use the absolute value test for convergence and root test it is inconclusive.

I was then thinking of using nth term test however I have the $(-1)^n$, can I just ignore that?

Any help would be greatly appreciated.

Hint: What can you say about $$\lim_{n\to\infty}\frac{(-1)^nn}{n+1}?$$

Look at $$\lim_{n\to\infty}\frac n{n+1}$$ first.

• ye I know that limit goes to one meaning the sum of the absolute series diverges, however that doesnt help since that doesn't prove that the alternating series converges or diverges, right? – user2250537 Jan 23 '16 at 17:56
• Well, since $\lim n/(n+1)=1$, $\lim (-1)^nn/(n+1)$ does not exist. – Tim Raczkowski Jan 23 '16 at 17:59
• the limit of any alternating series does not exist right, and if it doesn't exist does that mean it 'always' diverges? – user2250537 Jan 23 '16 at 18:00
• $\lim (-1)^n/n=0$. – Tim Raczkowski Jan 23 '16 at 18:02
• oh ok thanks so much! – user2250537 Jan 23 '16 at 18:03

As ${(-1)^nn\over n+1}\not\to 0$ the series does not converge.

• Oh so I can apply the nth term test to a series by just ignoring the $(-1)^n$ ? – user2250537 Jan 23 '16 at 17:59
• Because if so then I understand why it diverges – user2250537 Jan 23 '16 at 17:59
• @user2250537, the necessary condition is $\lim a_n = 0$. If this is false, maybe $\exists\lim a_n = l\ne 0$, maybe $\not\exists\lim a_n$, – Martín-Blas Pérez Pinilla Jan 23 '16 at 18:03

Fact. Let $\{a_n\}$ be a sequence. Then $\displaystyle\lim_{n\to\infty}a_n=0$ if and only if $\displaystyle\lim_{n\to\infty}\left\lvert a_n\right\rvert=0$

Now, we wish to investigate the convergence of $\sum a_n$ where $\displaystyle a_n=(-1)^n\frac{n}{n+1}$.

To do so, let's apply the Divergence Test.

Note that $$\lim_{n\to\infty}\left\lvert a_n\right\rvert = \lim_{n\to\infty}\left\lvert (-1)^n\frac{n}{n+1}\right\rvert = \lim_{n\to\infty}\frac{n}{n+1} = 1$$ It follows that $\displaystyle\lim_{n\to\infty}a_n\neq0$. Hence $\sum a_n$ diverges.