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I saw this problem:

Find $L=\lim_{n \to \infty } \sum\limits_{k=1}^n \frac{k^n}{n^n} $

Solutions to this limit can be found here.

I got $L=1.5$ which is wrong but I don't know why & where I went wrong.
We should have got $L=e/(e-1)$

My attempt :

By Faulhaber's formula $\displaystyle \sum_{k\le n} k ^r = \frac{n^{r+1}}{r+1}+ \frac{n^r}{2} + P_{n-1}{(r)} $ where $P_m(x) $ is a polynomial of $m$ degree.

So $$L=\lim_{n \to \infty }\sum_{k\le n}\frac{k^n}{n^n}= \lim_{n \to \infty }\frac{n}{n+1}+ \frac{n^n}{2n^n}+ \frac{P_{n-1}(n)}{n^n} =1.5$$

Since all the coefficients in Faulhaber's formula has an absolute value less than $1$ the last term should tend to $0$ because

$$|\frac{P_{n-1}(n)}{n^n}|< \sum_{k\ge1}\frac{1}{n^k} \to 0$$

Prem
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pie
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  • One problem is that $\frac{P_{n-1}(n)}{n^n}$ does not necessarily converge to zero. – Martin R Mar 05 '24 at 08:30
  • why? the biggest term in this polynomial is $n^{n-1}$ – pie Mar 05 '24 at 08:31
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    You cannot estimate $P_{n-1}(n) = O(n^{n-1})$ because the coefficients of $P_{n-1}$ can grow with $n$ as well. – Martin R Mar 05 '24 at 08:32
  • @MartinR I thought all the coefficients in Faulhaber's formula has an absolute value less than 1 – pie Mar 05 '24 at 08:35
  • There is a denominator in your formula, that is not in the Faulhaber's formula - how do you use it? – zkutch Mar 05 '24 at 08:36
  • @zkutch do you mean that $\frac{1}{2}$?, or what do you mean exactly ? – pie Mar 05 '24 at 08:36
  • $n^n$ is a constant so I took it outside of the sum – pie Mar 05 '24 at 08:39
  • Yes, thats correct. Seems argument about coefficients is right. – zkutch Mar 05 '24 at 08:40
  • @pie: I do not know if the coefficients are bounded, but even then you can not estimate like that because the number of terms increases. Btw, $r^n/2$ should probably be $n^r/2$. – Martin R Mar 05 '24 at 08:53
  • @MartinR if the coefficients are bounded it is easy to prove that the limit of $\frac{P_{n-1}(n)}{n^n}$ is less than $\sum_{k\ge1}\frac{1}{n^k}\to 0$, so that must mean coefficients are unbounded – pie Mar 05 '24 at 08:57
  • Faulhaber's formula gives $\sum_{k\le n}\frac{k^n}{n^n} = \sum_{r=0}^n \binom{n+1}{r} \frac{B_r}{(n+1)n^{r-1}}$, and all terms in that sum are $O(1)$. – Martin R Mar 05 '24 at 09:08
  • ... or better: All terms in that sum are of the form $B_n O(1)$. Still I do not see a reason why that “remainder” should converge to zero. – Martin R Mar 05 '24 at 09:25

2 Answers2

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Naturally , there is some mistake in the argument to get the Contradictory $L$ value.

Issue is here :
$\displaystyle \sum_{k\le n} k ^r = \frac{n^{r+1}}{r+1}+ \frac{n^r}{2} + \color{red}{P_{n-1}{(r)}}$

It should be :
$\displaystyle \sum_{k\le n} k ^r = \frac{n^{r+1}}{r+1}+ \frac{n^r}{2} + \color{blue}{P_{r-1}{(n)}}$

What this seemingly minor change does :

Earlier , where we had $\color{red}{P_{n-1}{(r)}}$ , the co-efficient of $r^{n-1}$ can not depend on $r$
... because then it implies that the term was not really $r^{n-1}$ & Polynomial is $r^n$ Degree.
Now , where we have $\color{blue}{P_{r-1}{(n)}}$ , the co-efficient of $n^{r-1}$ can (& does !!) depend on $r$ ,
... yet that term will remain really $n^{r-1}$ & Polynomial is indeed $n^{r-1}$ Degree.

Now , the co-efficient of $n^{r-1}$ is not Constant : Depending on $(r-1)$ , that co-efficient will vary , increasing or decreasing or whatever.

In a way , when we make $r=n$ , that co-efficient will then make the term $C n \times n^{n-1} / n^n = C$ & other terms will tend to $0$.

$$L=\lim_{n \to \infty }\sum_{k\le n}\frac{k^n}{n^n}= \lim_{n \to \infty }\frac{n}{n+1}+ \frac{n^n}{2n^n}+ \color{green}{\frac{P_{n-1}(n)}{n^n}} =1.5+\color{orange}{C}$$

The $C$ value will be suitable to make $L=e/(e-1)$

Prem
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1

I'll quote a section of the German Wikipedia about Faulhaber's formula (the English version seems not say anything similar to this)

Die niedrigen Koeffizienten in Stammbrüchen, wie man sie bei kleinem $k$ aus der Schulmathematik kennt, sind aber für den weiteren Verlauf überhaupt nicht typisch. Bereits bei $k = 11$ tritt zum ersten Mal ein Koeffizient > 1 auf; bei noch höheren Potenzen wird das zur Regel. Grund dafür sind die Bernoulli-Zahlen, die nach einer Reihe von niedrigen Werten stark ansteigen, sogar stärker als jede Exponentialfunktion, und gegen Unendlich gehen.

That translates (slightly shortened) to

The small coefficients, as known from school math for small values of $k$, are not typical for higher $k$. Already at $k=11$ comes the first coefficient bigger than $1$, for higher powers that becomes the norm. The reason are the Bernoulli numbers which, after some small values initially, increase significantly (even faster than any exponential function) and tend to infinity.

So your error lies in assuming that the coefficients of your $P_{n-1}(n)$ stay bounded so your argument that $\lim_{n \to \infty} \frac{P_{n-1}(n)}{n^n} = 0$ is incorrect.

Ingix
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  • When the coefficient is unbounded , then it must vary with $r$ , then the term was not really $r^{n-1}$ : It will go to $r \times r^{n-1} = r^n$ Degree which is not true ... Details in my answer ... – Prem Mar 05 '24 at 12:45