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

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?

• What's a "null sequence"? I'm not familiar with the term. – JonathanZ supports MonicaC May 20 '19 at 16:57
• Presumably a sequence tending to $0$. – 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.
• If you prove that the limit is $0$ then you shall have proved that the sequence is a null sequence. – José Carlos Santos May 20 '19 at 17:22
• 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. – ParabolicAlcoholic May 20 '19 at 17:23