# [Intro analysis] Prove lim s_n = 0

• fishturtle1
In summary: I'm not sure why you're bringing up ##a^{n-N}##. The goal is to show that for ##n > N##, ##\vert s_n \vert < a^{n-N} \vert s_N \vert##. You've only found that for ##n > N##, ##\vert s_n \vert < a \vert s_N \vert##.

## 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## ).

## 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 ...

You don't have to consider the cases for that hint. Notice that the hint says that ##L < 1##, so case 1 is sufficient.

Math_QED said:
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

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##. ...

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fishturtle1 said:
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)

fishturtle1 said:
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?

tnich said:
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.

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fishturtle1 said:
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##, . . .

tnich said:
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##

fishturtle1 said:
..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##.

fishturtle1
tnich said:
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

## What does lim s_n = 0 mean?

The notation lim s_n = 0 refers to the limit of a sequence, where s_n represents the nth term of the sequence. This means that as the value of n becomes larger and larger, the terms of the sequence approach the value of 0.

## How do you prove that lim s_n = 0?

To prove that lim s_n = 0, you must show that for any epsilon (ε) greater than 0, there exists a natural number N such that for all n > N, the absolute value of s_n - 0 is less than ε. This can be done using the epsilon-delta definition of a limit or by using other limit theorems and properties.

## Why is proving lim s_n = 0 important?

Proving that lim s_n = 0 is important because it confirms that the sequence converges to the value of 0. This is useful in many mathematical and scientific fields where understanding the behavior of sequences is important, such as in the study of functions, series, and calculus.

## What are some common techniques used to prove lim s_n = 0?

There are several techniques that can be used to prove that lim s_n = 0. These include the squeeze theorem, the comparison test, the Cauchy criterion, and the ratio test. Other methods may also be used depending on the specific characteristics of the sequence.

## Can lim s_n = 0 be proven for all sequences?

No, lim s_n = 0 cannot be proven for all sequences. Some sequences may not converge to the value of 0, and in those cases, the limit would be different. Additionally, some sequences may not have a limit at all, in which case the statement lim s_n = 0 would not be applicable.