Convergence of Sequences: Exploring the Limit Inside

[evinda]

Hello! (Wave)

I want to check the convergence of the sequences $\left( \left( 1+\frac{1}{\sqrt{n}}\right)^n\right)$, $\left( \left( 1+\frac{1}{2n}\right)^n\right)$.

We know that $e^x=\lim_{n \to +\infty} \left( 1+\frac{x}{n}\right)^n$.

We have that $\lim_{n \to +\infty} \left( 1+\frac{1}{\sqrt{n}}\right)^n=\lim_{n \to +\infty} \left( \left( 1+\frac{1}{\sqrt{n}}\right)^{\sqrt{n}}\right)^{\sqrt{n}}$.

Is the latter equal to $\lim_{n \to +\infty} e^{\sqrt{n}}=+\infty$ ?

So does it hold that $\lim_{n \to +\infty} \left( \left( 1+\frac{1}{\sqrt{n}}\right)^{\sqrt{n}}\right)^{\sqrt{n}}=\lim_{n \to +\infty} \left( \lim_{n \to +\infty}\left( 1+\frac{1}{\sqrt{n}}\right)^{\sqrt{n}}\right)^{\sqrt{n}} $ ?

If so, why? (Thinking)

 

[I like Serena]

Hello! (Wave)

I want to check the convergence of the sequences $\left( \left( 1+\frac{1}{\sqrt{n}}\right)^n\right)$, $\left( \left( 1+\frac{1}{2n}\right)^n\right)$.

We know that $e^x=\lim_{n \to +\infty} \left( 1+\frac{x}{n}\right)^n$.

We have that $\lim_{n \to +\infty} \left( 1+\frac{1}{\sqrt{n}}\right)^n=\lim_{n \to +\infty} \left( \left( 1+\frac{1}{\sqrt{n}}\right)^{\sqrt{n}}\right)^{\sqrt{n}}$.

Is the latter equal to $\lim_{n \to +\infty} e^{\sqrt{n}}=+\infty$ ?

So does it hold that $\lim_{n \to +\infty} \left( \left( 1+\frac{1}{\sqrt{n}}\right)^{\sqrt{n}}\right)^{\sqrt{n}}=\lim_{n \to +\infty} \left( \lim_{n \to +\infty}\left( 1+\frac{1}{\sqrt{n}}\right)^{\sqrt{n}}\right)^{\sqrt{n}} $ ?

If so, why? (Thinking)

Hey evinda!

I'm afraid that we cannot generally make such a deduction.
Consider for instance that:
$$e=\lim (1+\frac 1n)^n \ne \lim\left(\lim (1+\frac 1n)\right)^n = 1$$

Instead we should go back to the definition of a limit.
From your formula for $e^x$ we have that:
$$\forall\varepsilon>0\ \exists N\ \forall n> N:\left|\left( 1+\frac{x}{n}\right)^n - e^x \right| <\varepsilon \quad\Rightarrow\quad \left( 1+\frac{x}{n}\right)^n > e^x-\varepsilon$$don't we? (Wondering)

If we pick $x=1,\ \varepsilon=e-2,\ n=\sqrt m$, this becomes:
$$\exists N\ \forall m> N^2:\left( 1+\frac{1}{\sqrt m}\right)^{\sqrt m} > 2\quad\Rightarrow\quad
\left( 1+\frac{1}{\sqrt m}\right)^{m} > 2^{\sqrt m}$$
Since the right side diverges to $+\infty$, so does the left side. (Thinking)

 

[evinda]

Hey evinda!

I'm afraid that we cannot generally make such a deduction.
Consider for instance that:
$$e=\lim (1+\frac 1n)^n \ne \lim\left(\lim (1+\frac 1n)\right)^n = 1$$

Instead we should go back to the definition of a limit.
From your formula for $e^x$ we have that:
$$\forall\varepsilon>0\ \exists N\ \forall n> N:\left( 1+\frac{x}{n}\right)^n > e^x-\varepsilon$$
don't we? (Wondering)

If we pick $x=1,\ \varepsilon=e-2,\ n=\sqrt m$, this becomes:
$$\exists N\ \forall m> N^2:\left( 1+\frac{1}{\sqrt m}\right)^{\sqrt m} > 2\quad\Rightarrow\quad
\left( 1+\frac{1}{\sqrt m}\right)^{m} > 2^{\sqrt m}$$
Since the right side diverges to $+\infty$, so does the left side. (Thinking)

I see... (Nod)

For the sequence $\left( 1+\frac{1}{2n}\right)^{2n}$, I have thought the following.

Let $c_n=\left( 1+\frac{1}{2n}\right)^{n}$.

We have that $c_n^2=\left( 1+\frac{1}{2n}\right)^{2n}$ and so $\lim_{n \to +\infty} c_n^2=e$, so $1 \leq c_n^2 \leq 4$, for $n$ large enough.

Then we have that $1 \leq c_n \leq 2$.

But this doesn't help us... How else can we deduce something about the convergence of the sequence? (Thinking)

 

[I like Serena]

I see... (Nod)

For the sequence $\left( 1+\frac{1}{2n}\right)^{2n}$, I have thought the following.

Let $c_n=\left( 1+\frac{1}{2n}\right)^{n}$.

We have that $c_n^2=\left( 1+\frac{1}{2n}\right)^{2n}$ and so $\lim_{n \to +\infty} c_n^2=e$, so $1 \leq c_n^2 \leq 4$, for $n$ large enough.

Then we have that $1 \leq c_n \leq 2$.

But this doesn't help us... How else can we deduce something about the convergence of the sequence? (Thinking)

How about writing it as:
$$\left( 1+\frac{1}{2n}\right)^{n} = \left( 1+\frac{\frac 12}{n}\right)^{n}$$
(Thinking)

 

[evinda]

How about writing it as:
$$\left( 1+\frac{1}{2n}\right)^{n} = \left( 1+\frac{\frac 12}{n}\right)^{n}$$
(Thinking)

Ah yes, and so we get that it is equal to $\sqrt{e}$, right? (Smile)

 

[I like Serena]

Ah yes, and so we get that it is equal to $\sqrt{e}$, right?

Yep. (Nod)

 

[evinda]

Yep. (Nod)

Nice, thank you... (Happy)