2
$\begingroup$

Determine whether the following series converges: $$\sum_{n=2}^\infty\frac{(-1)^n}{\sqrt{n}-(-1)^{[\sqrt{n}]}}$$ where $$[x]=\max\{k\in\mathbb{Z}: k\leq x\}$$

My attempt: First I write $$\frac{(-1)^n}{\sqrt{n}-(-1)^{[\sqrt{n}]}}=\frac{(-1)^n(\sqrt{n}+(-1)^{[\sqrt{n}]})}{n-1}=\frac{(-1)^n\sqrt{n}}{n-1}+\frac{(-1)^{n+[\sqrt{n}]}}{n-1}$$

It is trivial that $\sum_{n=2}^\infty \frac{(-1)^n\sqrt{n}}{n-1}$ converges. Now I claim that the series $\sum_{n=2}^\infty\frac{(-1)^{n+[\sqrt{n}]}}{n-1}$ converges. Consider the sequence $n+[\sqrt{n}]$, this sequence is(starting from $n=1$): $$2,3,4,6,7,8,9,10,12,13,14,15,16,17,18,20,21\dots$$ we see that the numbers do not show up in the sequence are $5,11,19,29,\dots$, i.e. numbers of the form $m^2+m-1$, and these are all odd numbers.

Define $$a_n=\begin{cases}\frac{(-1)^{n+[\sqrt{n}]}}{n-1}, &n\text{ is not a complete squre}\\ 0, &\text{otherwise}\end{cases}$$

Then by alternating series test $\sum_{n=2}^\infty a_n$ converges. On the other hand, define $$b_n=\begin{cases}\frac{(-1)^{n+[\sqrt{n}]}}{n-1}=\frac{1}{n-1}, &n\text{ is a complete squre}\\ 0, &\text{otherwise}\end{cases}$$ then the series $\sum_{n=2}^\infty b_n$ is definitely convergent.

Observe that

$$\frac{(-1)^{n+[\sqrt{n}]}}{n-1}=a_n+b_n,\quad\forall n\in\mathbb{N}$$

therefore the series $\sum_{n=2}^\infty\frac{(-1)^{n+[\sqrt{n}]}}{n-1}$ is convergent. Hence the original series is convergent.

Does this proof look good?

$\endgroup$
  • $\begingroup$ The (true) fact that the omitted numbers are precisely $m^2+m-1$ should be proved, not "guessed" from the firs tfew terms. Of course $n+\lfloor \dqrt n\rfloor$ grows in steps of $1$, except when $\lfloor\sqrt n\rfloor $jumps - which is precisely at those values $\endgroup$ – Hagen von Eitzen Apr 16 '15 at 1:36
  • $\begingroup$ @HagenvonEitzen That's a good point. But I think this should not be very hard to justify. $\endgroup$ – Frank Lu Apr 16 '15 at 1:40

Your Answer

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

Browse other questions tagged or ask your own question.