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

  • #1
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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?
 

Answers and Replies

  • #2
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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.
 
  • #3
1,462
44
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?
 
  • #4
14,415
11,728
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.
 
  • #5
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? :)
 
  • #6
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.
 
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