How to prove that $\left(\frac{(-1)^n}{\sqrt{n}}\right)_{n\geq 1}$ is a null sequence?

My attempt via induction:

If I prove that the denominator grows faster than the numerator, I can conclude that it is indeed a null sequence, right? So I have to show that $\sqrt{n}\geq (-1)^n$ for $\forall n \in \mathbb{N}:n\geq 1$

Base case with $n_0=1$: $\sqrt{1}=1\geq (-1)^1=-1 \quad \checkmark$

Induction hypothesis: $\exists n\in \mathbb{N}:\sqrt{n}\geq (-1)^n$

Induction claim: $\Longrightarrow \sqrt{n+1} \geq (-1)^{n+1}$

Inductive step: $$\begin{gather}\sqrt{n+1}\geq (-1)^{n+1} \quad |\cdot (-1) \\ -\sqrt{n+1}\leq (-1)^{n+2} \end{gather}$$ This is true, since $-\sqrt{n+1}$ can't be $\geq -1$.

Is that a valid proof?

  • $\begingroup$ What's a "null sequence"? I'm not familiar with the term. $\endgroup$ – JonathanZ supports MonicaC May 20 '19 at 16:57
  • $\begingroup$ Presumably a sequence tending to $0$. $\endgroup$ – Alekos Robotis May 20 '19 at 16:57

You also have $n+1\geqslant n$ for each $n$, but $\left(\frac n{n+1}\right)_{n\in\mathbb N}$ is not a null sequence.

Note that $(\forall n\in\mathbb N):\left\lvert\frac{(-1)^n}{\sqrt n}\right\rvert=\frac1{\sqrt n}$ and that $\lim_{n\to\infty}\frac1{\sqrt n}=0$. This proves that your sequence is a null sequence.

  • $\begingroup$ So I just have to find the limit? $\endgroup$ – ParabolicAlcoholic May 20 '19 at 17:20
  • $\begingroup$ If you prove that the limit is $0$ then you shall have proved that the sequence is a null sequence. $\endgroup$ – José Carlos Santos May 20 '19 at 17:22
  • $\begingroup$ Let's get a little crazy. How would I then prove the $\left(\frac{n!}{n^n}\right)_{n \geq 0}$ is a null sequence. L'Hopitals Rule hasn't been taught yet. $\endgroup$ – ParabolicAlcoholic May 20 '19 at 17:23
  • $\begingroup$ That's another question. Please post it as such. $\endgroup$ – José Carlos Santos May 20 '19 at 17:24
  • $\begingroup$ Will do. Thanks. $\endgroup$ – ParabolicAlcoholic May 20 '19 at 17:25

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