# How do you determine whether the sequence a_n=n(-1)^n converges, if so how do you find the limit?

Mar 2, 2017

The sequence diverges.

#### Explanation:

We can apply the ratio test for sequences:

Suppose that;

$L = {\lim}_{n \rightarrow \infty} | {a}_{n + 1} / {a}_{n} | < 1 \implies {\lim}_{n \rightarrow \infty} {a}_{n} = 0$

i.e. if the absolute value of the ratio of successive terms in a sequence $\left\{{a}_{n}\right\}$ approaches a limit $L$, and if $L < 1$, then the sequence itself converges to $0$. It is important to note that this is a statement about the convergence of the sequence $\left\{{a}_{n}\right\}$, and it is not a statement about the series $\sum {a}_{n}$.

So for our sequence;

${a}_{n} = n {\left(- 1\right)}^{n}$

So our test limit is:

$L = {\lim}_{n \rightarrow \infty} | \frac{\left(n + 1\right) {\left(- 1\right)}^{n + 1}}{n {\left(- 1\right)}^{n}} |$
$\setminus \setminus \setminus = {\lim}_{n \rightarrow \infty} | \frac{\left(n + 1\right) {\left(- 1\right)}^{n} \left(- 1\right)}{n {\left(- 1\right)}^{n}} |$
$\setminus \setminus \setminus = {\lim}_{n \rightarrow \infty} | \frac{\left(n + 1\right) \left(- 1\right)}{n} |$
$\setminus \setminus \setminus = {\lim}_{n \rightarrow \infty} | \frac{\left(n + 1\right)}{n} |$
$\setminus \setminus \setminus = {\lim}_{n \rightarrow \infty} | 1 + \frac{1}{n} |$
$\setminus \setminus \setminus > 1$

And so the sequence does not converge.