# Math Help - Limit of sequence

1. ## Limit of sequence

Consider the sequence $f_{0}:N^{*}->R$
$f_{0}(n)=(1+\frac{1}{n})^{n}$
and define,
$f_{k}(n)=(\frac{f_{k-1}(n)}{\lim_{p\to\infty}f_{k-1}(p)})^{n}$.
n,k positive natural numbers.
Find:
$\lim_{n\to\infty}f_{k}(n)$

Using induction I found that
$f_{k}(n)=\frac{(1+\frac{1}{n})^{n^{k+1}}}{e^{\sum_ {i=1}^{k}(-1)^{i+1}\frac{n^{k+i-1}}{i}}}$
But I don't know how to find the limit.

The limit should be $e^{\frac{(-1)^{k}}{k+1}}$

2. Originally Posted by m3th0dman
Consider the sequence $f_{0}:N^{*}->R$
$f_{0}(n)=(1+\frac{1}{n})^{n}$
and define,
$f_{k}(n)=(\frac{f_{k-1}(n)}{\lim_{p\to\infty}f_{k-1}(p)})^{n}$.
n,k positive natural numbers.
Find:
$\lim_{n\to\infty}f_{k}(n)$

Using induction I found that
$f_{k}(n)=\frac{(1+\frac{1}{n})^{n^{k+1}}}{e^{\sum_ {i=1}^{k}(-1)^{i+1}\frac{n^{k+i-1}}{i}}}$
But I don't know how to find the limit.

The limit should be $e^{\frac{(-1)^{k}}{k+1}}$

According to what you wrote:

$f_1(n)=\left(\frac{f_0(n)}{\lim\limits_{n\to\infty }f_0(n)}\right)^n=\left(\frac{\left(1+\frac{1}{n}\ right)^n}{e}\right)^n$ $\xrightarrow [n\to\infty]{}1$ , so:

$f_2(n)=\left(\frac{f_1(n)}{\lim\limits_{p\to\infty }f_1(p)}\right)^n=
\left(\frac{\left(1+\frac{1}{n}\right)^n}{e}\right )^n=f_1(n)$
and etc...., so I think you may have not written what you meant...

Tonio

3. Originally Posted by tonio
According to what you wrote:

$f_1(n)=\left(\frac{f_0(n)}{\lim\limits_{n\to\infty }f_0(n)}\right)^n=\left(\frac{\left(1+\frac{1}{n}\ right)^n}{e}\right)^n$ $\xrightarrow [n\to\infty]{}1$ , so:

$f_2(n)=\left(\frac{f_1(n)}{\lim\limits_{p\to\infty }f_1(p)}\right)^n=
\left(\frac{\left(1+\frac{1}{n}\right)^n}{e}\right )^n=f_1(n)$
and etc...., so I think you may have not written what you meant...

Tonio
The following proposition is false:
$\lim_{n\to\infty}1^{n}=1$

So your assumption is not true.

$\lim_{n\to\infty}(\frac{(1+\frac{1}{n})^n}{e})^n=e ^{-\frac{1}{2}}$

4. Originally Posted by m3th0dman
The following proposition is false:
$\lim_{n\to\infty}1^{n}=1$

In fact this is trivially true. What is not true in general is that $f(n)\xrightarrow [n\to\infty]{}1\Longrightarrow f(n)^n\xrightarrow [n\to\infty]{}1$

So your assumption is not true.

No assumption but sheer misreading and consequent mistake of mine...

$\lim_{n\to\infty}(\frac{(1+\frac{1}{n})^n}{e})^n=e ^{-\frac{1}{2}}$

Indeed. And this limit is pretty hard to evaluate: I had to use twice L'Hospital' rule with logarithmic notation...

Tonio

5. Originally Posted by tonio
Indeed. And this limit is pretty hard to evaluate: I had to use twice L'Hospital' rule with logarithmic notation...

Tonio
I have been taught to avoid L'Hospital' rule, so here is what I wrote:

$\left(\frac{\left(1+\frac{1}{n}\right)^n}{e}\right )^n = e^{n(n\log(1+\frac{1}{n})-1)}$ $=e^{n(n(\frac{1}{n}-\frac{1}{2n^2}+o(\frac{1}{n^2}))-1)}=e^{-\frac{1}{2}+o(1)},$

and that's it.

Originally Posted by m3th0dman
Consider the sequence $f_{0}:N^{*}->R$
$f_{0}(n)=(1+\frac{1}{n})^{n}$
and define,
$f_{k}(n)=(\frac{f_{k-1}(n)}{\lim_{p\to\infty}f_{k-1}(p)})^{n}$.
n,k positive natural numbers.
Find:
$\lim_{n\to\infty}f_{k}(n)$
Moreover, the same method works for this general problem. Let us write $g_k(n)=\log f_k(n)$ for simplicity, and $\ell_k=\lim_n g_k(n)$, for all $k$.

Let $k\geq 1$. Start from:

$\log\left(1+\frac{1}{n}\right)=\frac{1}{n}-\frac{1}{2n^2}+\cdots+\frac{(-1)^{k+1}}{k n^k}+o\left(\frac{1}{n^k}\right)$.

The definition of $\ell_k$ is just an intricate way to describe the coefficients in this expansion.

Indeed,
$g_0(n)=n\log(1+\frac{1}{n})=1-\frac{1}{2n}+\cdots+\frac{(-1)^{k+1}}{k n^{k-1}}+o\left(\frac{1}{n^{k-1}}\right)=1+o(1)$,
hence $\ell_0=1$,
$g_1(n)=n(g_0(n)-\ell_0)=n(g_0(n)-1)=-\frac{1}{2}+\frac{1}{3n}+\cdots+\frac{(-1)^{k+1}}{k n^{k-2}}+o\left(\frac{1}{n^{k-2}}\right)=-\frac{1}{2}+o(1)$,
hence $\ell_1=-\frac{1}{2}$,
and so on, "unfolding the expansion", until
$g_{k-1}(n)=n(g_{k-2}(n)-\ell_{k-2})=\frac{(-1)^{k+1}}{k}+o(1)$,
hence $\ell_{k-1}=\frac{(-1)^{k+1}}{k}$.

If you really want to write every word of the proof, then you have to explicitate the above induction. Let $k\geq 1$ and prove by induction on $j=0,\ldots,k-1$ that :
$g_j(n)=\frac{(-1)^{j}}{j+1}+\frac{(-1)^{j+1}}{j+2}\frac{1}{n}+\cdots+\frac{(-1)^{k+1}}{k}\frac{1}{n^{k-j-1}}+o\left(\frac{1}{n^{k-j-1}}\right)$
(hence $\ell_j=\frac{(-1)^{j}}{j+1}$).

(you could also use convergent series instead of asymptotic expansions to shorten the proof, but these would be a little inappropriate here since we only care for the limit)

6. ## Thanks

Thank you very much!