# How would you prove this converges? $\sum_{1}^{\infty} \frac{(-1)^n}{[n-(-1)^n]^{\frac{2}{3}}}$

How would you check if this converges or not?

$$\sum_{1}^{\infty} \frac{(-1)^n}{[n-(-1)^n]^{\frac{2}{3}}}$$

It looks like a telescopic sequence so I thought I'd first write the beginning values:

$$S_n=\frac{-1}{2^{\frac{2}{3}}} + \frac{1}{1^{\frac{2}{3}}} + \frac{-1}{4^{\frac{2}{3}}} ...$$

But I got no conclusions from it... What am I missing here?

Thanks.

• Do you know the alternating serie test ? – Sylvain L. Jun 20 '15 at 17:22
• I know that I should write down the elements to see if they cancel each other out and see with what I've left with but it's not straightforward here – FigureItOut Jun 20 '15 at 17:24
• limit of integral as $n$ goes to infinity? – miniparser Jun 20 '15 at 17:25
• I don't think things will cancel nicely. But you can use the fact that $a_1+a_2+a_3+a_4+\cdots$ converges iff $a_2+a_1+a_4+a_3+\cdots$ converges and the Alternating Series Test. – David Mitra Jun 20 '15 at 17:32
• @SylvainL.: Despite your comment receiving two votes, the alternate series test does not work because $\frac 1 { {\sqrt {n - (-1)^n}} ^2}$ does not decrease. – Alex M. Jun 20 '15 at 21:09

We have: $$\frac{1}{\left[n+(-1)^n\right]^{\frac{2}{3}}} = \frac{1}{n^{\frac{2}{3}}}+O\left(\frac{1}{n^{\frac{5}{3}}}\right)$$ where: $$\sum_{n\geq 1}\frac{(-1)^n}{n^{5/3}}$$ is an absolutely convergent series and $$\sum_{n\geq 1}\frac{(-1)^n}{n^{2/3}}$$ is conditionally convergent by Leibniz' test.

You need two things:

(1) The terms tend to zero (though not monotonically).

(2) If you consider pairs of terms for $2n-1$ and $2n$,

if $f(n) = \frac1{n^{\frac{2}{3}}}$

$-\frac1{(2n)^{\frac{2}{3}}} +\frac1{(2n-1)^{\frac{2}{3}}} =-f(2n)+f(2n-1) =f'(x)$ where $2n-1 < x < 2n$ by the mean value theorem.

But $f'(x) = \frac23\frac1{x^{5/3}}$, so $\sum_{n=2}^{\infty} (-f(2n)+f(2n-1))$ converges.

• This solution should come with a cautionary tale for students. There's a slight issue with whether the particular rearrangement of this conditionally convergent series converges to the same thing (in general it doesn't, but here it works out), and you'd probably be expected to explicitly invoke the comparison test (in part because the value of $x$ depends on $n$). It'd be really easy to try to claim absolute convergence from this comparison test, which would be incorrect; that's part of the subtle issue with rearranging series. – zibadawa timmy Jun 24 '15 at 4:31

First, note that your series is not absolutely convergent: placing a modulus inside the sum you get $\sum \frac 1 {\big( n - (-1)^n \big)^{\frac 2 3}}$ which, keeping only the dominant term and applying the limit comparison test, behaves like $\sum \frac 1 {n^{\frac 2 3}}$ which is clearly divergent.

Now, is your series convergent? One cannot apply Leibniz's test because the sequence $\frac 1 {\big( n - (-1)^n \big)^{\frac 2 3}}$ does not decrease, even though it tends to $0$.

Let us examine once more the definition of the concept of convergent series: $x_1 + x_2 + \cdots + x_n + \cdots$ converges if and only if the sequences $(x_1 + x_2) + (x_3 + x_4) + \cdots + (x_{2N-1} + x_{2N})$ and $(x_1 + x_2) + (x_3 + x_4) + \cdots + (x_{2N-1} + x_{2N}) + x_{2N+1}$ converge, both, to the same limit, when $N \to \infty$.

In our case, $x_{2n+1} \to 0$, so we just have to show that $(x_1 + x_2) + (x_3 + x_4) + \cdots + (x_{2N-1} + x_{2N}) = \sum \limits _{n=1} ^N (x_{2n-1} + x_{2n})$ converges. In order to study this convergence we shall transform the summand as follows:

$$x_{2n-1} + x_{2n} = \\ -\frac 1 { {\sqrt {2n-1 +1}} ^2} + \frac 1 { {\sqrt {2n-1}} ^2} = \\ \Big( \frac 1 {\sqrt {2n-1}} + \frac 1 {\sqrt {2n}} \Big) \Big( \frac 1 {\sqrt {2n-1}} - \frac 1 {\sqrt {2n}} \Big) \sim \\ \frac 1 {\sqrt n} \frac {\sqrt {2n} - \sqrt {2n-1}} {\sqrt {2n} \sqrt {2n-1}} \sim \\ \frac 1 n \frac {\sqrt {2n} - \sqrt {2n-1}} 1 = \\ \frac 1 n \frac {2n - (2n-1)} { {\sqrt {2n}}^2 + {\sqrt {2n}} {\sqrt {2n-1}} + {\sqrt {2n-1}}^2 } \sim \\ \frac 1 {n^{\frac 5 3}} ,$$

where the notation $a_n \sim b_n$ means $a_n, b_n >0$ and $\lim \limits _{n \to \infty} \frac {a_n} {b_n} \in (0, \infty)$.

Since $\sum \frac 1 {n^{\frac 5 3}}$ converges, the convergence of your series follows now easily.

• The terms do not decrease. They do go to zero. If $n$ is even, $n-(-1)^n > n+1-(-1)^{n+1}$. – marty cohen Jun 20 '15 at 19:08
• @martycohen: Thank you for catching that, I've been superficial. I have changed my proof. – Alex M. Jun 20 '15 at 21:10

Let $(S_n)$ be the sequence of partial sums for the given series,

and let $(T_n)$ be the sequence of partial sums for $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^{2/3}}$.

Since this series converges by the Alternating Series Test, $\;T_n\to l$ for some number $l$

and therefore $S_{2n}=T_{2n}\to l$. $\;\;$Then $S_{2n+1}=S_{2n}-\frac{1}{(2n+2)^{2/3}}\to l$ also,

so $S_n\to l$ and the given series converges.