Another limit

Alexmahone

Active member
Find $\displaystyle \lim_{n\to\infty}\frac{(\ln n)^n}{[\ln (n+1)]^{n+1}}$.

chisigma

Well-known member
Find $\displaystyle \lim_{n\to\infty}\frac{(\ln n)^n}{[\ln (n+1)]^{n+1}}$.
Is...

$\displaystyle \frac{\ln n}{\ln (n+1)} = 1+ \frac{\ln n - \ln (n+1)}{\ln (n+1)} \implies \lim_{n \rightarrow \infty} (\frac{\ln n}{\ln (n+1)})^{n}=1 \implies$

$\displaystyle \implies \lim_{n \rightarrow \infty} \frac{1}{\ln (n+1)}\ (\frac{\ln n}{\ln (n+1)})^{n}=0$

Kind regards

$\chi$ $\sigma$

Plato

Well-known member
MHB Math Helper
Find $\displaystyle \lim_{n\to\infty}\frac{(\ln n)^n}{[\ln (n+1)]^{n+1}}$.
$\displaystyle\frac{(\ln n)^n}{[\ln (n+1)]^{n+1}}=\frac{1}{\ln(n+1}\left[\frac{\ln(n)}{\ln(n+1)}\right]^n$.

If $n\ge 3$ then $\displaystyle\left[\frac{\ln(n)}{\ln(n+1)}\right]<1$.

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Alexmahone

Active member
$\displaystyle\frac{(\ln n)^n}{[\ln (n+1)]^{n+1}}\frac{1}{\ln(n+1}\left[\frac{\ln(n)}{\ln(n+1)}\right]^n$.

If $n\ge 3$ then $\displaystyle\left[\frac{\ln(n)}{\ln(n+1)}\right]<1$.
Actually, $\displaystyle\left[\frac{\ln(n)}{\ln(n+1)}\right]<1$ for $n\ge 1$.

Plato

Well-known member
MHB Math Helper
Actually, $\displaystyle\left[\frac{\ln(n)}{\ln(n+1)}\right]<1$ for $n\ge 1$.
How does that effect the truth of what I posted?

Also sprach Zarathustra

Member
Find $\displaystyle \lim_{n\to\infty}\frac{(\ln n)^n}{[\ln (n+1)]^{n+1}}$.

Let us define the following infinite sum:

$$\sum_{n=1}^{\infty} \frac{(\ln n)^n}{[\ln (n+1)]^{n+1}}$$

That sum is converges, use Cauchy root test, hence,

$$\lim_{n\to\infty}\frac{(\ln n)^n}{[\ln (n+1)]^{n+1}}=0$$

Evgeny.Makarov

Well-known member
MHB Math Scholar
$\displaystyle \frac{\ln n}{\ln (n+1)} = 1+ \frac{\ln n - \ln (n+1)}{\ln (n+1)} \implies \lim_{n \rightarrow \infty} (\frac{\ln n}{\ln (n+1)})^{n}=1 \implies$

$\displaystyle \implies \lim_{n \rightarrow \infty} \frac{1}{\ln (n+1)}\ (\frac{\ln n}{\ln (n+1)})^{n}=0$
How did you conclude that $\displaystyle\lim_{n\to\infty} \left(1+ \frac{\ln n - \ln (n+1)}{\ln (n+1)}\right)^n=1$?

Alexmahone

Active member
How does that effect the truth of what I posted?
It doesn't, but you didn't complete your proof. $\displaystyle 1^\infty$ is an indeterminate form.

Evgeny.Makarov

Well-known member
MHB Math Scholar
Let us define the following infinite sum:

$$\sum_{n=1}^{\infty} \frac{(\ln n)^n}{[\ln (n+1)]^{n+1}}$$

That sum is converges, use Cauchy root test
If I am not mistaken, $\sqrt[n]{\frac{(\ln n)^n}{[\ln (n+1)]^{n+1}}}\to1$ as $n\to\infty$, so the test is inconclusive.

Plato

Well-known member
MHB Math Helper
It doesn't, but you didn't complete your proof. $\displaystyle 1^\infty$ is an indeterminate form.
There is nothing to complete. You have a bounded factor and a factor the limit of which is $=?$

Alexmahone

Active member
There is nothing to complete. You have a bounded factor and a factor the limit of which is $=?$
zero

chisigma

Well-known member
How did you conclude that $\displaystyle\lim_{n\to\infty} \left(1+ \frac{\ln n - \ln (n+1)}{\ln (n+1)}\right)^n=1$?
Because is...

$\displaystyle (1+o(n))^{n}= 1 + n\ o(n) + \binom {n}{2}\ o^{2}(n)+...+ \binom{n}{n-1}\ o^{n-1}(n) + o^{n}(n)$ (1)

... You have that...

$\displaystyle \lim_{n \rightarrow \infty} n\ o(n)=0 \implies \lim_{n \rightarrow \infty}(1+o(n))^{n}=1$ (2)

For n 'large enough' is...

$\displaystyle \ln (n+1)-\ln n \sim \frac{1}{n} \implies o(n)= \frac{\ln n - \ln (n+1)}{\ln (n+1)} \sim - \frac{1}{n\ \ln (n+1)}$ (3)

... so that the (2) is verified...

Kind regards

$\chi$ $\sigma$

Evgeny.Makarov

Well-known member
MHB Math Scholar
$\displaystyle \lim_{n \rightarrow \infty} n\ o(n)=0 \implies \lim_{n \rightarrow \infty}(1+o(n))^{n}=1$
By the definition of small-o, $f(n)$ is $o(n)$ if $f(n)/n\to 0$, not $nf(n)\to0$. Also, $1/n$ is $o(1)$ and $o(n)$, but $(1+1/n)^n\to e$ as $n\to\infty$.