alexmahone
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Find $\displaystyle \lim_{n\to\infty}\frac{(\ln n)^n}{[\ln (n+1)]^{n+1}}$.
Alexmahone said:Find $\displaystyle \lim_{n\to\infty}\frac{(\ln n)^n}{[\ln (n+1)]^{n+1}}$.
$\displaystyle\frac{(\ln n)^n}{[\ln (n+1)]^{n+1}}=\frac{1}{\ln(n+1}\left[\frac{\ln(n)}{\ln(n+1)}\right]^n$.Alexmahone said:Find $\displaystyle \lim_{n\to\infty}\frac{(\ln n)^n}{[\ln (n+1)]^{n+1}}$.
Plato said:$\displaystyle\frac{(\ln n)^n}{[\ln (n+1)]^{n+1}}\frac{1}{\ln(n+1}\left[\frac{\ln(n)}{\ln(n+1)}\right]^n$.
If $n\ge 3$ then $\displaystyle\left[\frac{\ln(n)}{\ln(n+1)}\right]<1$.
How does that effect the truth of what I posted?Alexmahone said:Actually, $\displaystyle\left[\frac{\ln(n)}{\ln(n+1)}\right]<1$ for $n\ge 1$.
Let us define the following infinite sum:Alexmahone said:Find $\displaystyle \lim_{n\to\infty}\frac{(\ln n)^n}{[\ln (n+1)]^{n+1}}$.
How did you conclude that $\displaystyle\lim_{n\to\infty} \left(1+ \frac{\ln n - \ln (n+1)}{\ln (n+1)}\right)^n=1$?chisigma said:$\displaystyle \frac{\ln n}{\ln (n+1)} = 1+ \frac{\ln n - \ln (n+1)}{\ln (n+1)} \implies \lim_{n \rightarrow \infty} (\frac{\ln n}{\ln (n+1)})^{n}=1 \implies$
$\displaystyle \implies \lim_{n \rightarrow \infty} \frac{1}{\ln (n+1)}\ (\frac{\ln n}{\ln (n+1)})^{n}=0$
Plato said:How does that effect the truth of what I posted?
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.Also sprach Zarathustra said: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
There is nothing to complete. You have a bounded factor and a factor the limit of which is $=?$Alexmahone said:It doesn't, but you didn't complete your proof. $\displaystyle 1^\infty$ is an indeterminate form.
Plato said:There is nothing to complete. You have a bounded factor and a factor the limit of which is $=?$
Evgeny.Makarov said:How did you conclude that $\displaystyle\lim_{n\to\infty} \left(1+ \frac{\ln n - \ln (n+1)}{\ln (n+1)}\right)^n=1$?
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 $n\to\infty$.chisigma said:$\displaystyle \lim_{n \rightarrow \infty} n\ o(n)=0 \implies \lim_{n \rightarrow \infty}(1+o(n))^{n}=1$