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This question stems from and old revision of this question, in which an upper bound for $n!$ was asked for.

The original bound was incorrect. In fact, I want to show that the given expression divided by $n!$ goes to $0$ as $n$ tends to $\infty$.

I thus want to show: $$\lim_{n\to\infty}\frac{(n+1)^{n^2+n+1}}{n!(n+2)^{n^2+1}}=0.$$

Using Stirling's approximation, I found that this is equivalent to showing that $$\lim_{n\to\infty} \frac{\exp(n)}{\sqrt n}\cdot\left(\frac{n+1}{n+2}\right)^{n^2+1}\cdot\left(\frac{n+1}{n}\right)^n=0.$$

However, I don't see how to prove the latter equation.

EDIT: It would already be enough to determine the limit of $$\exp(n)\left(\frac{n+1}{n+2}\right)^{(n^2)}\left(\frac{n+1}{n}\right)^n$$ as $n$ goes to $\infty$.

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    $\begingroup$ use log and exp to solve it. $\endgroup$ – Pallavi Jun 4 at 16:21
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    $\begingroup$ can you divide by n in the brackets and factor the n outside, then partly cancel, leaving $n^n$ on the top - and err, work some other stuff out to do with e and e^n and stuff $\endgroup$ – Cato Jun 4 at 16:31
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    $\begingroup$ @CheerfulParsnip the limit approaches 0 graphically $\endgroup$ – Jake Freeman Jun 4 at 16:33
  • $\begingroup$ How detailed do we want to be? We can use the Calc I method of just getting rid of the $1$ and the $2$ in the binomials, because reasons. $\endgroup$ – The Count Jun 4 at 16:39
  • $\begingroup$ @JakeFreeman, my mistake. I carelessly assumed that $((n+1)/(n+2))^{n^2+1}$ approached $1$, which was a dumb mistake for me to make. I've deleted my comment. $\endgroup$ – Cheerful Parsnip Jun 4 at 16:56
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The easy part first: We have by basic analysis (where $x\in\Bbb R$) \begin{equation}\tag 1\label 1\lim_{x\to\infty} \left(\frac{x+1}x\right)^x=\lim_{x\to\infty}(1+1/x)^x=e.\end{equation}

Now comes the harder part:
Note that \begin{equation}\label 2\tag 2 \lim_{x\to\infty} e^x \left(\frac{x+1}{x+2}\right)^{(x^2)} = \lim_{x\to\infty} \exp\left(x+x^2\ln\left({x+1\over x+2}\right)\right). \end{equation}

We now have by Taylor expansion of $\ln(1-y)$ (for $x$ big enough): \begin{align}\tag 3\label 3 x+x^2 \ln(1-\frac1{x+2}) &=x-\sum_{k=1}^\infty\frac1k\frac{x^2}{(x+2)^k} \\ &=x-\frac{x^2}{x+2}-\frac{x^2}{2(x^2+4x+4)}-\overbrace{\sum_{k=3}^\infty \frac{x^2}{k(x+2)^k}}^{\xrightarrow{x\to\infty} 0}. \end{align}

The latter sum converges to $0$ since (for $x> 1$), $\displaystyle\sum_{k=3}^\infty \frac{x^2}{k(x+2)^k}\le\sum_{k=3}^\infty \frac{x^2}{x^k}=\sum_{k=1}^\infty x^{-k}=\frac1x\frac{x}{x-1}=\frac1{x-1}$.

Thus, by additivity of the limit, \begin{align}\label 4\tag 4 \lim_{x\to\infty} x+x^2\ln(1-\frac1{x+2}) = \lim_{x\to\infty} \overbrace{x-\frac{x^2}{x+2}}^2-\overbrace{\frac{x^2}{2(x^2+4x+4)}}^\frac12=\frac32. \end{align}

We can now use continuity of the exponential function and \eqref{2} to find that \begin{align}\tag 5\label 5 \lim_{x\to\infty} e^x \left(\frac{x+1}{x+2}\right)^{(x^2)} &= \exp\left(\lim_{x\to\infty} x+x^2\ln\left({x+1\over x+2}\right)\right) \\ &= e^{3/2}.&\eqref 4 \end{align}

We can thus finally assert, using multiplicity of the limit, that your limit equals $0$:

\begin{equation} \bbox[5px,border:2px solid #C0A000]{ \lim_{x\to\infty} \color{orange}{\frac{x+1}{(x+2)\sqrt x}} \color{blue}{e^x \left(\frac{x+1}{x+2}\right)^{(x^2)}} \color{green}{\left(\frac{x+1}x\right)^x} = \color{orange}0 \cdot \color{blue}{e^{3/2}} \cdot \color{green}e = 0. } \end{equation}

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    $\begingroup$ I like the colors. $\endgroup$ – Spencer Kraisler Jun 4 at 23:54
  • $\begingroup$ Very concise! Thanks $\endgroup$ – user679638 Jun 4 at 23:57
  • $\begingroup$ MUCH PRETTY. Also the answer is very clear. Thank you! By the way, here was the question I actually care about: math.stackexchange.com/questions/3250980/… But this proof here is AWESOME! The answer that was mentioned on this linked question doesn't actually answer the question (proving a limit is a certain value isn't as strong as deriving a strong bound (since the latter can usually also arrive to the former)). $\endgroup$ – Squirtle Jun 5 at 2:46
  • $\begingroup$ @Squirtle For your original question, see here: math.stackexchange.com/q/2564744/631742 $\endgroup$ – Maximilian Janisch Jun 5 at 8:43
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$$\frac{(n+1)^{n^2+n+1}}{n! (n+2)^{n^2+1}}$$

= $$\frac{(1+\frac{1}{n})^{n^2+n+1}}{n! (1+\frac{2}{n})^{n^2+1}} \frac{n^{n^2+n+1}}{n^{n^2+1}}$$

=$$\frac{(1+\frac{1}{n})^{n^2+n+1}}{n! (1+\frac{2}{n})^{n^2+1}} \frac{n^nn^{n^2+1}}{n^{n^2+1}}$$

=$$\frac{(1+\frac{1}{n})^{n^2+n+1}}{n! (1+\frac{2}{n})^{n^2+1}} n^n$$

=$$\frac{((1+\frac{1}{n})^{n})^n(1+\frac{1}{n})^{n}(1+\frac{1}{n})}{n! ((1+\frac{2}{n})^{n})^n (1+\frac{2}{n}) } n^n$$

at the limit n tends to infinity, using the standard definition of log(z) = $\lim_{x\to\infty}(1 + 1/z)^z$

=$$\frac{e^n.e.1}{n!e^{2n}e^2} n^n$$

=$$\frac{ n^n}{n!e^{n+2}} $$

is how far I got really

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