/sci/ - Science & Math » Thread #12717563

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Anonymous Wed 17 Feb 21:05:03 2021 No.12717563 View Reply Original Report

Quoted By: >>12717671 >>12717813 >>12717847 >>12718054 >>12718162 >>12718616

Theorem: $\lim_{n \rightarrow \infty} a^n = 0$ for all $a \neq 1$ .

Proof:
$ \begin{align} L &= \lim_{n \rightarrow \infty} a^n \\ &= \lim_{n \rightarrow \infty} a^{n+1} \\ &= \lim_{n \rightarrow \infty} a a^n \\ &= a \lim_{n \rightarrow \infty} a^n \\ &= a L \\ L - a L &= 0 \\ (1 - a) L &= 0 \\ 1 - a &\neq 0 \\ L &= 0 \end{align} $
QED

Anonymous

Anonymous Wed 17 Feb 2021 21:21:41 No.12717671 Report

Quoted By: >>12717775

>>12717563
sixth row is true only if L is a real number

Anonymous

Anonymous Wed 17 Feb 2021 21:42:57 No.12717775 Report

Quoted By:

>>12717671
Incorrect.
Consider $L = i$ .

Anonymous

Anonymous Wed 17 Feb 2021 21:47:26 No.12717799 Report

Quoted By: >>12717826

I think if you subtract a*L on both sides of
L=a*L
you're assuming your conclusion

Anonymous

Anonymous Wed 17 Feb 2021 21:53:26 No.12717813 Report

Quoted By: >>12717819

>>12717563
Since everything after the first line is false in the limit how is that a proof?

Anonymous

Anonymous Wed 17 Feb 2021 21:54:48 No.12717819 Report

Quoted By: >>12718005 >>12719132

>>12717813
> $\lim_{n \rightarrow \infty} f(n) \neq \lim_{n \rightarrow \infty} f(n+1)$

Anonymous

Anonymous Wed 17 Feb 2021 21:56:06 No.12717826 Report

Quoted By:

>>12717799
Nope. $A = B \Rightarrow f(A) = f(B)$ for any $f$ .

Anonymous

Anonymous Wed 17 Feb 2021 22:01:37 No.12717847 Report

Quoted By: >>12717918

>>12717563
At the very moment you give a name to the limit, you assume it exists as a real number. Assuming the limit exists your reasoning is actually correct. Obviously it's not always the case that it converges, you need |a| < 1 for that.

Anonymous

Anonymous Wed 17 Feb 2021 22:24:59 No.12717918 Report

Quoted By: >>12717974

>>12717847
>At the very moment you give a name to the limit, you assume it exists as a real number.
nope, L can be ? as long as the performed operations make sense.

Anonymous

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Anonymous Wed 17 Feb 2021 22:38:18 No.12717974 Report

Quoted By: >>12718033

>>12717918
You've been quite literally filtered, brainlet.

Anonymous

Anonymous Wed 17 Feb 2021 22:48:23 No.12718005 Report

Quoted By: >>12718009 >>12718025 >>12718562

>>12717819
This is the answer. Likewise

$\lim_{n \rightarrow \infty} x neq \lim_{n \rightarrow \infty} x^2, \qquad x \neq x^2$

Anonymous

Anonymous Wed 17 Feb 2021 22:49:35 No.12718009 Report

Quoted By:

>>12718005
4chan doesn't let me delete posts for some reason.

$\lim_{n \rightarrow \infty} x \eq \lim_{n \rightarrow \infty} x^2, \qquad x \neq x^2$

Anonymous

Anonymous Wed 17 Feb 2021 22:52:15 No.12718025 Report

Quoted By:

>>12718005
Are you a retard? If a sequence converges to a value, any subsequence of it converges to the same value. This thread is not even worth my time.

Anonymous

Anonymous Wed 17 Feb 2021 22:53:57 No.12718033 Report

Quoted By:

>>12717974
>I have no argument

Anonymous

Anonymous Wed 17 Feb 2021 23:00:52 No.12718054 Report

Quoted By: >>12718060 >>12718069

>>12717563
This is wrong. Consider for example, the following.

$L = \sum_{i=1}^\infty \frac{1}{2^i} = 1$ but that $\sum_{i=1}^\infty \frac{1}{2^{i+1}} = \frac{1}{2} \neq L$

Anonymous

Anonymous Wed 17 Feb 2021 23:02:54 No.12718060 Report

Quoted By: >>12718064 >>12718221

>>12718054
It's not a series, it's a fucking limit.

Anonymous

Anonymous Wed 17 Feb 2021 23:03:39 No.12718064 Report

Quoted By: >>12718078

>>12718060
>implying series are limits

Anonymous

Anonymous Wed 17 Feb 2021 23:05:57 No.12718069 Report

Quoted By:

>>12718054
you have done something entirely different

Anonymous

Anonymous Wed 17 Feb 2021 23:06:49 No.12718078 Report

Quoted By:

>>12718064
A subseries does not converge to the value of the source series. Worse, if you don't have absolute convergence, reordering the terms of a series can produce an arbitrary result. Please go back to high school.

Anonymous

Anonymous Wed 17 Feb 2021 23:10:28 No.12718093 Report

Quoted By: >>12718099

It does not hold when $a > 1$
because when $a > 1, L - aL \neq 0$

Anonymous

Anonymous Wed 17 Feb 2021 23:11:38 No.12718099 Report

Quoted By: >>12718567

>>12718093
Yeah it only works when a = 0, otherwise you can't divide by L to get L = 0.

Anonymous

Anonymous Wed 17 Feb 2021 23:29:11 No.12718162 Report

Quoted By: >>12718181

>>12717563
You have failed to prove such series converges. You cannot assume a limit converges and then perform manipulations on it, otherwise you get nonsense. Watch:

Theorem: $\lim_{n \rightarrow \infty} (1+n) = 0$

Proof:

$\begin{align} L &= \lim_{n \rightarrow \infty} (1+n) \\ &= \lim_{n \rightarrow \infty} (1+(2n+1)) \\ &= \lim_{n \rightarrow \infty} ((n+1) + (n+1)) \\ &= \lim_{n \rightarrow \infty} (n+1) + \lim_{n \rightarrow \infty} (n+1)\\ &= L + L \\ L &= L + L \\ L &= 0 \end{align}$

QED

Anonymous

Anonymous Wed 17 Feb 2021 23:34:20 No.12718181 Report

Quoted By: >>12718195

>>12718162
all true besides the ultimate line

Anonymous

Anonymous Wed 17 Feb 2021 23:37:15 No.12718195 Report

Quoted By:

>>12718181
Last line must be true, because it's convergent.

Anonymous

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Anonymous Wed 17 Feb 2021 23:44:16 No.12718221 Report

Quoted By: >>12718576

>>12718060
are series not a limit of partial sums of a sequence?

Anonymous

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Anonymous Thu 18 Feb 2021 01:28:43 No.12718562 Report

Quoted By:

>>12718005
>he doesn't understand greentext

Anonymous

Anonymous Thu 18 Feb 2021 01:30:23 No.12718567 Report

Quoted By:

>>12718099
There's no division anywhere retard

Anonymous

Anonymous Thu 18 Feb 2021 01:31:23 No.12718576 Report

Quoted By:

>>12718221
Yeah but your manipulation doesn't translate

Anonymous

Anonymous Thu 18 Feb 2021 01:42:39 No.12718616 Report

Quoted By: >>12719129

>>12717563
Lim a^n
=
Lim (m!-c)
=
Infinity-c
=
Infinity

Anonymous

Anonymous Thu 18 Feb 2021 03:42:39 No.12718976 Report

Quoted By: >>12719132

Making the substitution a^n -> a^(n+1) is only valid if, in the limit of large n, a^(n+1)-a^n approaches zero, which it doesn’t if a>1

Anonymous

Anonymous Thu 18 Feb 2021 04:49:12 No.12719129 Report

Quoted By:

>>12718616
Elaborate

Anonymous

Anonymous Thu 18 Feb 2021 04:50:14 No.12719132 Report

Quoted By:

>>12718976
No dummy, see >>12717819
Limits are invariant to shifting

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