6
$\begingroup$

I'm having a lot of trouble figuring this one out.

Determine if $\sum_{i=2}^{\infty} \frac{(-1)^n}{n+(-1)^n}$ converges or diverges

Both ratio and root test are inconclusive and I'm at a loss. Can anyone help me?

$\endgroup$
  • $\begingroup$ Have you tried the alternating series test too? $\endgroup$ – R_D Oct 25 '15 at 6:41
  • $\begingroup$ Combine successive pairs of terms, and use the fact that the limit of the original terms is $0$ to show that the partial sums converge. $\endgroup$ – André Nicolas Oct 25 '15 at 6:50
  • $\begingroup$ @AndréNicolas Oups, I didn't see your comment, it seems we have the same idea :-) $\endgroup$ – Jean-Claude Arbaut Oct 25 '15 at 9:58
  • $\begingroup$ @Jean-ClaudeArbaut: Your solution is well-explained. $\endgroup$ – André Nicolas Oct 25 '15 at 10:03
2
$\begingroup$

Let

$$S_n=\sum_{k=2}^n \frac{(-1)^n}{n+(-1)^n}$$

Take the sum of two consecutive terms:

$$\frac{(-1)^{2n}}{2n+(-1)^{2n}}+\frac{(-1)^{2n+1}}{2n+1+(-1)^{2n+1}}=\frac{1}{2n+1}-\frac{1}{2n}=\frac{-1}{2n(2n+1)}$$

Thus $S_{2n+1}$ converges. Since the terms tend to $0$, $S_{2n}$ is also converging, to the same limit.

$\endgroup$
1
$\begingroup$

One may observe that, as $n \to \infty$, we have $$ \begin{align} \frac{(-1)^n}{n+(-1)^n}&=\frac{(-1)^n}{n}\times\frac1{1+\frac{(-1)^n}{n}}\\\\ &=\frac{(-1)^n}{n}\left(1-\frac{(-1)^n}{n}+\mathcal{O}\left(\frac1{n^2} \right)\right)\\\\ &=\frac{(-1)^n}{n}-\frac1{n^2}+\mathcal{O}\left(\frac1{n^3} \right) \end{align} $$ then, for some integer $p$,

$$ \underbrace{\sum_{n\geq p}\frac{(-1)^n}{n+(-1)^n}}_{{\color{red}{\text{conditionally CV}}}}=\underbrace{\sum_{n\geq p}\frac{(-1)^n}{n}}_{{\color{red}{\text{conditionally CV}}}}-\underbrace{\sum_{n\geq p}\frac1{n^2}+\sum_{n\geq p}\mathcal{O}\left(\frac1{n^3} \right)}_{{\color{blue}{\text{absolutely CV}}}} $$

and your initial series is conditionally convergent.

$\endgroup$
0
$\begingroup$

Let's write

$$\frac{(-1)^n}{n+(-1)^n} = \left ( \frac{(-1)^n}{n+(-1)^n} - \frac{(-1)^n} {n} \right) + \frac{(-1)^n}{n}.$$

The term in parentheses equals $-1/[(n+(-1)^n)n].$ In absolute value, these terms are $\le 1/(n-1)^2.$ Hence the series of these terms converges absolutely. The series $\sum (-1)^n/n$ converges by the alternating series test. So our series is the sum of two convergent series, hence converges.

$\endgroup$
-1
$\begingroup$

Terms are alternatively positive and negative, decrease in absolute value and converge to $0$, so the sseries converges (this is the alternating series test), but it does not converge absolutely.

$\endgroup$
  • 7
    $\begingroup$ The terms do not decrease in absolute value. $\endgroup$ – zhw. Oct 25 '15 at 6:45
  • 1
    $\begingroup$ It's amazing that this blatantly wrong answer got upvoted twice. People should check before upvoting... $\endgroup$ – Najib Idrissi Oct 25 '15 at 8:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.