Proof of a convergent sequence

[muted remnant]

For the tip you may start with:

$\(a-b) \cdot (a^{n-1} + a^{n-2}b + a{n-3}b^2 + ... + ab^{n-2} + b^{n-1})\$

If you multiply the whole thing by a and then by -b and then sort it, you will notice that everything cancels nicely out.

[coral sundial]

adonhs

[muted remnant]

I am currently thinking about this too

Suppose if

$\ \lim_{i\to\infty} \sqrt[n]{a_i} = \sqrt[n]{\lim_{i\to\infty} {a_i}}
\Rightarrow \lim_{i\to\infty} \sqrt[n]{a_i} - \sqrt[n]{ \lim_{i\to\infty} {a_i} } = 0$

[coral sundial]

adonhs

[muted remnant]

And maybe we can use the identity now?

Rewrite it as:

$\ \lim_{i\to\infty} (a_i)^{\frac{1}{n}} - (\lim_{i\to\infty} \ {a_i})^{\frac{1}{n}} = 0$

[coral sundial]

adonhs

[muted remnant]

finally

No I just moved it to left side

And maybe you can show with this that the stuff inside the right parentheses equals to 0 maybe..

Write the following maybe as:

$\ (\lim_{i\to\infty} (a_i) - (\lim_{i\to\infty} {a_i})) \cdot \(\lim_{i\to\infty} (a_i)^{\frac{1}{n} - 1} + \lim_{i\to\infty} (a_i)^{\frac{1}{n} - 2} \cdot (\lim_{i\to\infty} {a_i})+\lim_{i\to\infty} (a_i)^{\frac{1}{n} - 3} \cdot (\lim_{i\to\infty} \ {a_i})^2 + ... + \lim_{i\to\infty} (a_i) \cdot (\lim_{i\to\infty} \ {a_i})^{\frac{1}{n} - 2} + (\lim_{i\to\infty}{a_i})^{\frac{1}{n} - 1})= 0$

[coral sundial]

Infinity

[muted remnant]

Is

$\ \lim_{i\to\infty} (a_i)^{\frac{1}{n}}\$

the same as

$\ (a_i)^{(\frac{1}{n})}\$

?

[coral sundial]

adonhs

[muted remnant]

is this the question?

[coral sundial]

adonhs

[muted remnant]

$\
\Rightarrow (\lim_{i\to\infty} (ai)^{\frac{1}{n} - 1} + \lim_{i\to\infty} (ai)^{\frac{1}{n} - 2} \cdot (\lim_{i\to\infty} {ai})+\lim_{i\to\infty} (ai)^{\frac{1}{n} - 3} \cdot (\lim_{i\to\infty} \ {ai})^2 + ... + \lim_{i\to\infty} (ai) \cdot (\lim_{i\to\infty} \ {ai})^{\frac{1}{n} - 2} + (\lim_{i\to\infty}{a_i})^{\frac{1}{n} - 1})= 0$

[coral sundial]

adonhs

[muted remnant]

How can I answer this questions I mean, in the second term you left the lim i-> inf out so to me it's like nothing saying

We know that $a_i$ is for all i convergent

[coral sundial]

adonhs

[muted remnant]

wait

[muted remnant]

Look at that equation

we have two products

If we consider this product:

$\ \lim_{i\to\infty} (a_i) - (\lim_{i\to\infty} {a_i}) = 0\$

question to you, is it true?

[coral sundial]

adonhs

[muted remnant]

The whole idea again:

$\$

Suppose if

$\ \lim_{i\to\infty} \sqrt[n]{a_i} = \sqrt[n]{\lim_{i\to\infty} {a_i}}
\Rightarrow \lim_{i\to\infty} \sqrt[n]{a_i} - \sqrt[n]{ \lim_{i\to\infty} {a_i} } = 0\$

Rewrite it as

$\ \lim_{i\to\infty} (a_i)^{\frac{1}{n}} - (\lim_{i\to\infty} \ {a_i})^{\frac{1}{n}} = 0\$

Use the identity

$\ (\lim_{i\to\infty} (a_i) - (\lim_{i\to\infty} {a_i})) \cdot ( ... ) = 0\$

Now we use the zero product property

$\ \lim_{i\to\infty} (a_i) - (\lim_{i\to\infty} {a_i}) = 0
\Rightarrow \lim_{i\to\infty} (ai) = (\lim_{i\to\infty} {a_i})$ ... which is true

[coral sundial]

adonhs

[muted remnant]

Say we have x² - x = 0

Then we can do $\x \cdot (x-1) = 0\$

This equation has two products.

The zero product property says if at least one product is 0 then the whole thing of products becomes 0.

[coral sundial]

adonhs

[muted remnant]

So by using the identity we are again left we two products in the identity.

If we take a look at the first product
$\ (\lim_{i\to\infty} (a_i) - (\lim_{i\to\infty} {a_i})) = 0$

[coral sundial]

adonhs

[muted remnant]

And the transform into
$\Rightarrow \lim_{i\to\infty} (a_i) = (\lim_{i\to\infty} {a_i})$

[coral sundial]

adonhs

[muted remnant]

This limit is true they are the same

Let's begin from there

The whole idea again:

$\$

Suppose if

$\ \lim_{i\to\infty} \sqrt[n]{a_i} = \sqrt[n]{\lim_{i\to\infty} {a_i}}
\Rightarrow \lim_{i\to\infty} \sqrt[n]{a_i} - \sqrt[n]{ \lim_{i\to\infty} {a_i} } = 0\$

Rewrite it as

$\ \lim_{i\to\infty} (a_i)^{\frac{1}{n}} - (\lim_{i\to\infty} \ {a_i})^{\frac{1}{n}} = 0\$

Use the identity

$\ (\lim_{i\to\infty} (a_i) - (\lim_{i\to\infty} {a_i})) \cdot ( ... ) = 0\$

Now we use the zero product property

$\ \lim_{i\to\infty} (a_i) - (\lim_{i\to\infty} {a_i}) = 0
\ \Rightarrow \lim_{i\to\infty} (ai) = (\lim_{i\to\infty} {a_i})$ ... which is true

[coral sundial]

adonhs

[muted remnant]

We can transform our previous assumption and try to prove that

Now we can use the identity that we just have proven before.

(...) it this long and messy stuff that I tried to write 1000 times

[muted remnant]

It's this

But apparently we don't need it anyway

ok wait

$\ \lim_{i\to\infty} (a_i) = (\lim_{i\to\infty} {a_i})$
\Rightarrow \lim_{i\to\infty} (a_i) - (\lim_{i\to\infty} {a_i}) = 0\$
Do you agree?

[coral sundial]

adonhs
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[muted remnant]

Because since $a_i$ is convergent this statement is true.

[coral sundial]

adonhs

[muted remnant]

for all i

You gotta understand this step

Ok now take a look at the identity, we have:

$\ (\lim_{i\to\infty} (a_i) - (\lim_{i\to\infty} {a_i})) \cdot ( ... ) = 0\$

[coral sundial]

adonhs

[muted remnant]

I wrote (...) which I refer to this long mess

Can you follow?

Ok you know now that
$\\lim_{i\to\infty} (a_i) - (\lim_{i\to\infty} {a_i}) = 0\$

[coral sundial]

adonhs

[muted remnant]

So what happens to this identity equation

If you substitude

Didn't I?

The whole idea again:

$\$

Suppose if

$\ \lim_{i\to\infty} \sqrt[n]{a_i} = \sqrt[n]{\lim_{i\to\infty} {a_i}}
\Rightarrow \lim_{i\to\infty} \sqrt[n]{a_i} - \sqrt[n]{ \lim_{i\to\infty} {a_i} } = 0\$

Rewrite it as and try to prove this since it's equivalent to the previous transformation

$\ \lim_{i\to\infty} (a_i)^{\frac{1}{n}} - (\lim_{i\to\infty} \ {a_i})^{\frac{1}{n}} = 0\$

Use the identity

$\ (\lim_{i\to\infty} (a_i) - (\lim_{i\to\infty} {a_i})) \cdot ( ... ) = 0\$

[coral sundial]

adonhs

[muted remnant]

And
$\ (...) = (\lim_{i\to\infty} (a_i)^{\frac{1}{n} - 1} + \lim_{i\to\infty} (a_i)^{\frac{1}{n} - 2} \cdot (\lim_{i\to\infty} {a_i})+\lim_{i\to\infty} (a_i)^{\frac{1}{n} - 3} \cdot (\lim_{i\to\infty} \ {a_i})^2 + ... + \lim_{i\to\infty} (a_i) \cdot (\lim_{i\to\infty} \ {a_i})^{\frac{1}{n} - 2} + (\lim_{i\to\infty}{a_i})^{\frac{1}{n} - 1})\$

which is pretty long and confusing, but I claim we don't need it anyway.

[coral sundial]

adonhs

[muted remnant]

If we prove that the identity equation equals 0

$\ (\lim_{i\to\infty} (a_i) - (\lim_{i\to\infty} {a_i})) \cdot ( ... ) = 0\$

Then by using the identity, we showed that the assumptions is true

[coral sundial]

adonhs

[muted remnant]

We know $a_i$ is convergent ∀ i ∈ $\mathbb{N}$

$

\ \lim_{i\to\infty} (a_i) - (\lim_{i\to\infty} {a_i}) = 0 \

\ \Leftrightarrow \lim_{i\to\infty} (a_i) = (\lim_{i\to\infty} {a_i}) \

\ \Leftrightarrow (\lim_{i\to\infty} (a_i) - (\lim_{i\to\infty} {a_i})) \cdot ( ... ) = 0$ ...Identity \

$ \Leftrightarrow \lim_{i\to\infty} (a_i)^{\frac{1}{n}} - (\lim_{i\to\infty} \ {a_i})^{\frac{1}{n}} = 0 \

\ \Leftrightarrow \lim_{i\to\infty} \sqrt[n]{a_i} - \sqrt[n]{ \lim_{i\to\infty} {a_i} } = 0 \

\ \Leftrightarrow \lim_{i\to\infty} \sqrt[n]{a_i} = \sqrt[n]{\lim_{i\to\infty} {a_i}}\$

[coral sundial]

adonhs
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[muted remnant]

Ok now I think it should be understandable, this was my idea/approach the whole time, how to make use of the identity. So i am only 90 % sure that this solid

Does it at least make sense?

Like if the 1 product equals to zero then 0 $\cdot$ anything = 0 and then further of that we can conclude the previous statement should also be equal to zero and if they are equal to zero then the limit should be the same

[coral sundial]

adonhs

[muted remnant]

Yeah so we can ignore the second product since the first is already fulfilling the role of being 0.

And it's zero (step 2) because in step (1) we say that the limits of those two are equal which they are if you read the task

[muted remnant]

Actually I should have not use equivalence arrows but still if you read the proof from below to the top it should make sense

Because reading it from the top to below isn't necessarily the cae

[muted remnant]

@nova sable Ok I just got confirmation that the identity only holds for positive integers

So we cannot yet use the identity instantly because instead of ${a_i}^n$ we got ${a_i}^{\frac{1}{n}}$

[coral sundial]

adonhs

[muted remnant]

This also why in the task it's says after proving the identity to apply it somehow to ${a_i}^{\frac{1}{n}}$ and ${a}^{\frac{1}{n}}$

[coral sundial]

adonhs

[muted remnant]

I think there is a way by defining:

$\a \coloneqq \sqrt[n]{a_i}$ and $b \coloneqq \sqrt[n]{\lim_{i\to\infty} a_i}\$

And then consider $a^n - b^n = 0\$

You will then have to use the identity, transform the equation and then use the limit operator on both sides and make a conclusion or so...