Showing that if lim a^2 = 0 implies lim a = 0

[Mr Davis 97]

Homework Statement

Suppose that $a_n$ is a sequence such that $\sum_{n=1}^\infty a_n^2$ converges. Show that $\lim_{n\to\infty}a_n = 0$.

Homework Equations

The Attempt at a Solution

My idea was this. Since $a_n^2$ converges, we have that $\lim_{n\to\infty}a_n^2 = 0$. I want to claim that $\lim_{n\to\infty}a_n^2 = (\lim_{n\to\infty}a_n)^2 = 0$, but is this justified? Don't I have to know that $a_n$ does in fact converge before I can use the algebra of limits?

 

[fresh_42]

You cannot reason that way: For $a_n=(-1)^n$ we have $\lim a_n^2=1$ whereas $\lim a_n$ does not exist. So at least more care is needed.

 

[Mr Davis 97]

You cannot reason that way: For $a_n=(-1)^n$ we have $\lim a_n^2=1$ whereas $\lim a_n$ does not exist. So at least more care is needed.

I think I can show it using an epsilon-delta proof:

Let $\epsilon > 0$. There exists $N$ such that $n\ge N$ implies that $|a_n^2| < \epsilon^2$. So $|a_n| < \epsilon$, by taking the square root of both sides. Is this correct? Is this the best way to show this?

 

[fresh_42]

I think I can show it using an epsilon-delta proof:

Let $\epsilon > 0$. There exists $N$ such that $n\ge N$ implies that $|a_n^2| < \epsilon^2$. So $|a_n| < \epsilon$, by taking the square root of both sides. Is this correct? Is this the best way to show this?

Yes.

You implicitly used $\sum |b_n| < \infty \Longrightarrow \lim |b_n|=0\,$, so I'm not sure if you have this or still need to show it.

 

[member 587159]

You cannot reason that way: For $a_n=(-1)^n$ we have $\lim a_n^2=1$ whereas $\lim a_n$ does not exist. So at least more care is needed.

Nice profile picture. What/who exactly are these people? :)

 

[member 587159]

Easy solution: $a_n^2 \to 0 \implies |a_n| \to 0 \implies a_n \to 0$

where I used that square root is continuous.