TITLE: Bounds of $1^n + 2^{n-1} + 3^{n-2} + \cdots + n^1$
QUESTION [17 upvotes]: I want to estimate the value of this sequence for large $n$ (with a reasonable lower bound and upper bound).
That is, can we find a function $f(n)$ such that $$ \frac{1^n + 2^{n-1} + 3^{n-2} + \cdots + n^1}{f(n)} \rightarrow 1?$$
I tried to approximate $ (n - x)^x $ as a Gaussian, look at how close it is visually, but the manipulation escapes me.
Here is the case where $n=20$:


Addendum: This post might be relevant.

REPLY [4 votes]: Starting from @Chrystomath's answer, for a given value of $a$ we have analyical expressions for $A$, $b$ and $c$.
The trick is to rewrite his/her second equation
$$(a-b)\ln(a-b)=b$$
$$(a-b)+(a-b)\ln(a-b)=a \implies k+k \log(k)=a \implies k=\frac{a}{W(e a)}$$ where $W(.)$ is Lambert function. So, this gives
$$\color{blue}{A=k^{a-k}\qquad b=a-k \qquad  c=\frac{2 k^2}{a+k}}\qquad \text{with} \qquad  \color{red}{k=\frac{a}{W(e a)}}$$ and then
$$A\sqrt{\pi c}= k^{a-k+1} \sqrt{\frac{2 \pi}{a+k}}$$ Rounding the numbers, for $a=20$,  this generates the sequence
$$\{2,4,9,23,66,210,733,2780,11374,49845,232672,1151412,6016082,33072048,190691716,
   1150116697,7238323772,47432585137,323004401255,2281724622065, 16693240814087\}$$ instead of
$$\{1,3,8,22,65,209,732,2780,11377,49863,232768,1151914,6018785,33087205,190780212,11
   50653920,7241710929,47454745803,323154696184, 2282779990494, 16700904488705\}$$
Edit
All the work can be done without expansions but instead function identification at a single point.
Consider
$$f(x)=(a-x)^x $$
$$f'(x)=(a-x)^x \left(\log (a-x)-\frac{x}{a-x}\right)$$
$$f''(x)=(a-x)^{x-2} \left((a-x) \log (a-x) ((a-x) \log (a-x)-2 x)-2 a+x^2+x\right)$$ $$ g(x)=A e^{-\frac{(x-b)^2}{c}}$$ $$g'(x)=-\frac{2 A (x-b) e^{-\frac{(x-b)^2}{c}}}{c}$$ $$ g''(x)=-\frac{2 A e^{-\frac{(b-x)^2}{c}} \left(c-2 (b-x)^2\right)}{c^2}$$ Compute $x_*$ corresponding to $f'(x)=0$ and then solve for $(A,b,c)$ the equations
$$f(x_*)=g(x_*) \qquad \qquad f'(x_*)=g'(x_*) \qquad \qquad f''(x_*)=g''(x_*) $$ For sure, this leads to the same results.