# [Intro analysis] Prove lim s_n = 0

## 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 ...

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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.

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, $\vert \frac{s_{n+1}}{s_n} \vert - \vert L \vert < \vert \frac{s_{N+1}}{s_N} - L \vert$. ...

Last edited:
tnich
Homework Helper
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?

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:
tnich
Homework Helper
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 $\vert s_{n} \vert < a \vert s_{n-1} \vert$, . . .

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 $n > N$. $\square$

tnich
Homework Helper
..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 $lim~s_n = 0$.

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