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[Intro analysis] Prove lim s_n = 0

  • #1
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Homework Statement


Assume all ##s_n \neq 0## and that the limit ##L = \lim \vert \frac{s_{n+1}}{s_n} \vert ## exists.

a) Show that if ##L < 1##, then ##\lim s_n = 0##. (Hint: Select ##a## so that ##L < a < 1## and obtain ##N## s.t. ##\vert s_{n+1} \vert < a \vert s_n \vert## for ##n \ge N##. Then show ##\vert s_n \vert < a^{n - N} \vert s_N \vert## for ##n > N## ).

Homework Equations


Limit laws for addition/multiplication/scalar

The Attempt at a Solution


Proof of hint:
Let ##\varepsilon_1 = \frac{1 - L}{2}## and select ##a = L + \varepsilon_1##.
Then ##L < a < 1##.

Consider 3 cases:

Case 1: ##L < 1##. Then ##\varepsilon_1 > 0##.
By definition, for all ##\varepsilon > 0## there exists ##N## such that ##n > N## implies ##\vert \frac{s_{n+1}}{s_n} - L \vert < \varepsilon##. By triangle inequality and substitution, we have ##\vert \frac{s_{n+1}}{s_n} \vert - \vert L \vert \le \vert \frac{s_{n+1}}{s_n} - L \vert < \varepsilon_1##. So ##\frac{\vert s_{n+1} \vert}{\vert s_n \vert} < \varepsilon_1 + \vert L \vert = \varepsilon_1 + L = a##. Multiplying both sides by ##\vert s_n \vert## we get ##\vert s_{n+1} \vert < \vert s_n \vert a##.

Case 2: ##L > 1##. Then ##\varepsilon_1 < 0##. ...

Case 3: ##L = 1##. Then ##\varepsilon_1 = 0## ... So now i can't substitute for ##\varepsilon > 0## in the limit definition ...

I am asking for advice on how to proceed, please.
 

Answers and Replies

  • #2
Math_QED
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You don't have to consider the cases for that hint. Notice that the hint says that ##L < 1##, so case 1 is sufficient.
 
  • #3
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You don't have to consider the cases for that hint. Notice that the hint says that ##L < 1##, so case 1 is sufficient.
! I completely missed that, thank you. I am going to start on the actual problem now then
 
  • #4
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well maybe I am still stuck on the hint,
edit:the case 1 is wrong i think.
Let ##L < 1##. Then ##\varepsilon_1 = \frac{1 - L}{2} > 0##. By definition, there exists ##N## such that ##n > N## implies ##\vert \frac{s_{n+1}}{s_n} - L \vert < \varepsilon_1##. By triangle inequality, ##\frac{\vert s_{n+1} \vert}{\vert s_n \vert} < \varepsilon_1 + \vert L \vert##. But ##L## could be less than 0. So I can't do the substitute ##a## like I did in OP.

Continuing from case 1...

We've shown ##\vert s_{n+1} \vert < a \vert s_n \vert##. We want to show ##\vert s_n \vert< a^{n-N} \vert s_N \vert## for ##n > N##.
For ##n > N##, ##\vert \frac{s_{n+1}}{s_n} - L \vert < \vert \frac{s_{N+1}}{s_N} - L \vert < \varepsilon##. By triangle inequality, ##\vert \frac{s_{n+1}}{s_n} \vert - \vert L \vert < \vert \frac{s_{N+1}}{s_N} - L \vert##. ...
 
Last edited:
  • #5
tnich
Homework Helper
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well maybe I am still stuck on the hint,
edit:the case 1 is wrong i think.
Let ##L < 1##. Then ##\varepsilon_1 = \frac{1 - L}{2} > 0##. By definition, there exists ##N## such that ##n > N## implies ##\vert \frac{s_{n+1}}{s_n} - L \vert < \varepsilon_1##. By triangle inequality, ##\frac{\vert s_{n+1} \vert}{\vert s_n \vert} < \varepsilon_1 + \vert L \vert##.
Not quite true. ##\left|\frac{\vert s_{n+1} \vert}{\vert s_n \vert}-\vert L \vert\right| \leq \left|\frac{\vert s_{n+1} \vert}{\vert s_n \vert}-L\right|##. (Edited)

But ##L## could be less than 0. So I can't do the substitute ##a## like I did in OP.

Continuing from case 1...

We've shown ##\vert s_{n+1} \vert < a \vert s_n \vert##. We want to show ##\vert s_n \vert< a^{n-N} \vert s_N \vert## for ##n > N##.
For ##n > N##, ##\vert \frac{s_{n+1}}{s_n} - L \vert < \vert \frac{s_{N+1}}{s_N} - L \vert < \varepsilon##. By triangle inequality, ##\vert \frac{s_{n+1}}{s_n} \vert - \vert L \vert < \vert \frac{s_{N+1}}{s_N} - L \vert##. ...
##L = \lim \vert \frac{s_{n+1}}{s_n} \vert## and ##\vert \frac{s_{n+1}}{s_n} \vert > 0 ~\forall n##. So how can ##L## be less than zero?
 
  • #6
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Not quite true. ##\left|\frac{\vert s_{n+1} \vert}{\vert s_n \vert}-\vert L \vert\right| \leq \left|\frac{\vert s_{n+1} \vert}{\vert s_n \vert}-L\right|##. (Edited)


##L = \lim \vert \frac{s_{n+1}}{s_n} \vert## and ##\vert \frac{s_{n+1}}{s_n} \vert > 0 ~\forall n##. So how can ##L## be less than zero?
Thanks for the reply, Here is what I have now from #5,

Proof of hint: Choose ##a = L + \frac{1-L}{2}##. We observe ##L \ge 0## since ##\vert\frac{s_{n+1}}{s_n}\vert > 0## for all ##n## so ##0## is always closer to our sequence than any negative number.

By definition of limit, there exists ##N## such that ##n \ge N## implies ##\frac{1-L}{2} > \vert \vert \frac{s_{n+1}}{s_n} \vert - L \vert \ge \vert \frac{s_{n+1}}{s_n} \vert - L## (since ##L \ge 0##). Adding ##L## and multiplying by ##\vert s_n \vert## gives ##\vert s_{n+1} \vert < \vert s_n \vert a##. ...

I'm not sure about the ##a^{n-N}##... It doesn't seem like ##(\vert s_n \vert)## is decr but the hint is suggesting that, I know that ##n - N > 0## so ##a^{N-n} < a## so ##\vert s_n \vert < a^{N-n}\vert s_n \vert##. Edit: I see that since ##L < 1##, ##(\vert s_n \vert)## must be str decr.

Edit2: Ok so from where we left off ... ##n > N## implies ## \vert s_{n+1} \vert < \vert s_n \vert a < \vert s_n\vert < \vert s_N \vert##. So ##\vert s_n \vert a < \vert s_N \vert##. Also, ##0 \le a^{n-N} \le a < 1##.

edit3: I think this is a proof by induction, will post when I have something.
 
Last edited:
  • #7
tnich
Homework Helper
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Thanks for the reply, Here is what I have now from #5,

Proof of hint: Choose ##a = L + \frac{1-L}{2}##. We observe ##L \ge 0## since ##\vert\frac{s_{n+1}}{s_n}\vert > 0## for all ##n## so ##0## is always closer to our sequence than any negative number.

By definition of limit, there exists ##N## such that ##n \ge N## implies ##\frac{1-L}{2} > \vert \vert \frac{s_{n+1}}{s_n} \vert - L \vert \ge \vert \frac{s_{n+1}}{s_n} \vert - L## (since ##L \ge 0##). Adding ##L## and multiplying by ##\vert s_n \vert## gives ##\vert s_{n+1} \vert < \vert s_n \vert a##. ...

I'm not sure about the ##a^{n-N}##... It doesn't seem like ##(\vert s_n \vert)## is decr but the hint is suggesting that, I know that ##n - N > 0## so ##a^{N-n} < a## so ##\vert s_n \vert < a^{N-n}\vert s_n \vert##. Edit: I see that since ##L < 1##, ##(\vert s_n \vert)## must be str decr.

Edit2: Ok so from where we left off ... ##n > N## implies ## \vert s_{n+1} \vert < \vert s_n \vert a < \vert s_n\vert < \vert s_N \vert##. So ##\vert s_n \vert a < \vert s_N \vert##. Also, ##0 \le a^{n-N} \le a < 1##.
If ##\vert s_{n+1} \vert < a \vert s_n \vert##, then ##\vert s_{n} \vert < a \vert s_{n-1} \vert##, . . .
 
  • #8
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If ##\vert s_{n+1} \vert < a \vert s_n \vert##, then ##\vert s_{n} \vert < a \vert s_{n-1} \vert##, . . .
..So ##\vert s_{n+1} \vert < a^2 \vert s_{n-1} \vert##.. I think that is what you are suggesting? I use this idea below,

Proof: We know ##a \vert s_n \vert > \vert s_{n+1} \vert## for all ##n \ge N##. We proceed by induction. Let ##P(n)## be the assertion that ##a^{n-N} \vert s_N \vert > s_n## for all ##n > N##.

Base case: If ##n = N + 1## we know that ##a^{N+1 - N} \vert s_{N} \vert = a \vert s_N \vert > \vert s_{N+1} \vert## is true.

Inductive step: Suppose ##P(n)## is true for some ##n > N##. Then ##a^{n-N}\vert s_N \vert > \vert s_n \vert##. We also know ##a \vert s_n \vert > \vert s_{n+1} \vert##. Combining these inequalities gives, ##a^{n+1 - N} \vert s_N \vert > a \vert s_n \vert > \vert s_{n+1} \vert##. Thus, ##a^{(n+1) - N} \vert s_N \vert > \vert s_{n+1} \vert##. So ##P(n+1)## is true.

We can conclude ##P(n)## is true for all ##n > N##. ##\square##
 
  • #9
tnich
Homework Helper
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..So ##\vert s_{n+1} \vert < a^2 \vert s_{n-1} \vert##.. I think that is what you are suggesting? I use this idea below,

Proof: We know ##a \vert s_n \vert > \vert s_{n+1} \vert## for all ##n \ge N##. We proceed by induction. Let ##P(n)## be the assertion that ##a^{n-N} \vert s_N \vert > s_n## for all ##n > N##.

Base case: If ##n = N + 1## we know that ##a^{N+1 - N} \vert s_{N} \vert = a \vert s_N \vert > \vert s_{N+1} \vert## is true.

Inductive step: Suppose ##P(n)## is true for some ##n > N##. Then ##a^{n-N}\vert s_N \vert > \vert s_n \vert##. We also know ##a \vert s_n \vert > \vert s_{n+1} \vert##. Combining these inequalities gives, ##a^{n+1 - N} \vert s_N \vert > a \vert s_n \vert > \vert s_{n+1} \vert##. Thus, ##a^{(n+1) - N} \vert s_N \vert > \vert s_{n+1} \vert##. So ##P(n+1)## is true.

We can conclude ##P(n)## is true for all ##n > N##. ##\square##
Well done. Now you just need one more step to show that ##lim~s_n = 0##.
 
  • #10
272
28
Well done. Now you just need one more step to show that ##lim~s_n = 0##.
Full proof,

Proof: Assume ##s_n \neq 0## for all ##n## and ##L = \lim \vert \frac{s_{n+1}}{s_n} \vert## exists.

Choose ##a = L + \frac{1 - L}{2}##. Then ##L < a < 1##. By def of limit, there exists ##N## such that ##n \ge N## implies ##\vert \vert \frac{s_{n+1}}{s_n} \vert - L \vert < \frac{1 - L}{2}##. Observe, ##\frac{1 - L}{2} > \vert \vert \frac{s_{n+1}}{s_n} \vert - L \vert \ge \vert \frac{s_{n+1}}{s_n} \vert - L## since ##L \ge 0##. So ##\vert \frac{s_{n+1}}{s_n} \vert < L + \frac{1-L}{2} = a##. So ##\vert s_{n+1} \vert < a\vert s_n\vert##.

Now let ##P(n)## be the assertion that ##\vert s_n \vert < a^{n-N}\vert s_N \vert## for all ##n > N##. We proceed by induction.

(Base case) We've shown that ##\vert s_{N+1} \vert < a^{N+1-N}\vert s_N \vert = a \vert s_N \vert##.

(Inductive step) Suppose ##P(n)## is true for some ##n > N##. Then ##\vert s_n\vert < a^{n-N}\vert s_N \vert## for some ##n > N##. We also know ##\vert s_{n+1} \vert < a \vert s_n \vert ##. So, ##\vert s_{n+1} \vert < a \vert s_n \vert < a^{n+1 - N} \vert s_N \vert##. Thus ##\vert s_{n+1} \vert < a^{(n+1) - N} \vert s_N \vert## that is ##P(n+1)## is true.

We can conclude ##P(n)## is true for all ##n > N##.

Finally, we use the Squeeze lemma to show ##\lim \vert s_n \vert = 0##. We know ##0 \le \vert s_n \vert \le a^{n}\cdot \frac{1}{a^N}\cdot\vert s_N \vert##. We know ##\lim 0 = 0##. Also, ##\lim a^n = 0## and ##\lim \frac{1}{a^N} = \frac{1}{a^N}## and ## \lim \vert s_N \vert = \vert s_N \vert##. So ##\lim (a^n\cdot \frac{1}{a^N} \cdot \vert s_N \vert) = 0 \cdot \frac{1}{a^N} \cdot \vert s_N \vert = 0##. By Squeeze lemma, we can conclude ##\lim \vert s_n \vert = 0##. ##\square##

Thank you for your help and guiding me through this
 

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