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I am evaluating the classical series $\lim\limits_{n\to\infty}\sum\limits_ {k=1}^{n-1}\frac{k^n}{n^n}$, but I want to know why the following solution of mine gives a wrong answer.

First I used the fact that $1^n+2^n+\cdots+k^n=f_n(k)$, where $f_n(k)$ is a polynomial with leading term $\frac1{n+1}k^{n+1}$. Thus, we have $$\lim\limits_{n\to\infty}\sum\limits_{k=1}^{n-1}\frac{k^n}{n^n}=\lim_{n\to\infty}\frac1{n^n}\sum_{k=1}^{n-1}k^n=\lim_{n\to\infty}\frac{\frac1{n+1}(n-1)^{n+1}}{n^n}=\lim_{n\to\infty}\left(1-\frac1n\right)^n\frac{n-1}{n+1}=\frac1{\rm e}.$$

The answer should be $\frac1{\rm e-1}$ though.

Bowei Tang
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youthdoo
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    why do you think that only the leading term in $f_n(n-1)$ contributes to the limit - if $n$ was fixed and $k$ much larger that would be a good guess but here $k=n-1$ so the following terms of lower degree in $k=n-1$ may contribute since their coefficients which also depend on $n$ may balance for the lower degree – Conrad Jun 12 '25 at 04:45
  • Can you ignore the term after the leading term in $f_n(k)$ (the term involving the $n$'th power)? The denominator has only the $n$'th power and so it is only terms involving powers below $n$ that can be ignored. – Jayanth R Varma Jun 12 '25 at 04:47
  • $\lim_{k\to\infty}\frac{f(k,n)}{g(k,n)}=1,,,\not!!!!\implies\lim_{n\to\infty}\frac{f(n-1,n)}{g(n-1,n)}=1$, just like how multivariable limits work – Quý Nhân Jun 12 '25 at 05:03
  • @Conrad By considering the rest of the terms, is it possible to fix the solution? – youthdoo Jun 12 '25 at 05:13
  • @QuýNhân I think this is indeed my reasoning, but I still find it difficult to understand the "$\not\Rightarrow$". – youthdoo Jun 12 '25 at 05:14
  • @youthdoo the first limit approaches $\infty$ along horizontal line, the second limit approaches $\infty$ along diagonal line, they are different paths. Limits with different paths do not have to be equal. – Quý Nhân Jun 12 '25 at 05:22
  • See https://math.stackexchange.com/q/4451257 – Gary Jun 12 '25 at 09:29

2 Answers2

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It's true that, as $k \rightarrow \infty$, $f_n(k) \sim k^{n+1}/(n+1)$, but this does not hold uniformly in $n$ and $k$. The issue is that a $n$ grows alongside $k$, it also increases the number of "large" terms, which ends up contributing non-trivially to the limit.

The robust way to take this limit is to make the substitution $j = n - k$, which renders \begin{align*} \sum_{k=1}^{n-1}\Bigl(\frac{k}{n}\Bigr)^n = \sum_{j=1}^{n-1}\Bigl(\frac{n - j}{n}\Bigr)^n = \sum_{j=1}^\infty \Bigl(1 - \frac{j}{n}\Bigr)^n \mathbf{1}_{\{j < n\}}. \end{align*}

The monotone convergence theorem then allows us to commute the limit with the sum so that \begin{align*} \lim_{n\rightarrow \infty} \sum_{k=1}^{n-1}\Bigl(\frac{k}{n}\Bigr)^n = \sum_{j=1}^\infty \lim_{n\rightarrow \infty}\Bigl(1 - \frac{j}{n}\Bigr)^n \mathbf{1}_{\{j < n\}} = \sum_{j=1}^\infty e^{-j} = \frac{1}{e-1}. \end{align*}

Damian Pavlyshyn
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In comments, you have been explained where is the error.

Using generalized harmonic numbers $$S_n=\sum\limits_{k=1}^n\frac{k^n}{n^n}=\frac{1}{n^n}\sum\limits_{k=1}^nk^n=\frac{1}{n^n}H_n^{(-n)}$$

If you look at sequence $A031971$ in $OEIS$, you will find two approximations $$ H_n^{(-n)}\sim \frac{e}{e-1}\,n^n$$ by Benoit Cloitre in year $2003$ and $$ H_n^{(-n)}\sim \frac{(e n+1)}{(e-1) (n+1)}\,n^n$$ by myself in year $2018$.

Both of them give the limit.

Edit

You could even show that $$S_n=\frac{e}{e-1}-\frac{1}{n}+\frac{1}{n^2}+O\left(\frac{1}{n^3}\right)$$

Claude Leibovici
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