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#1 h Mar 29, 2025, 7:42 AM

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Find: $\lim_{n \to \infty} n\cdot\sum_{k=1}^n \frac 1 {k(n-k)!}$

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#2 h Mar 29, 2025, 8:25 AM • 2 Y

Y by Snoop76, Svyatoslav

Rewriting gives $\sum_{k=0}^{n-1} {n\over {(n-k)k!}}$ . Now define $f_n(k)={n\over {(n-k)k!}}$ if $k<n$ and $f_n(k)=0$ otherwise. Then the sum equals $\sum_0^\infty f_n(k)$ which is basically an integral with counting measure. We have $f_n(0)=1$ and when $n\ge 2,k\ge 1$ we have $0\le f_n(k)\le {2\over {(k-1)!}}$ , which is summable.
Also $f_n(k)\rightarrow 1/k!$ pointwise. Hence we can apply Lebesgue and find $\sum_0^\infty f_n(k)\rightarrow \sum_0^\infty 1/k!=e$ .
We can avoid Lebesgue and use a $\delta/\epsilon$ proof, if we like.

Thank you Solyaris for the correction.

This post has been edited 1 time. Last edited by alexheinis, Mar 31, 2025, 9:41 AM

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#3 h Mar 31, 2025, 7:31 AM

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A minor issue: In your notation $f_n(k) > \frac 1 {k!}$ . But this is easily fixed:

$f_n(0) = 1$ and for $k \ge 1$ we have $\frac 1 {(n-k)k} \le \frac 1 {n-1}$ and thus $f_n(k) \le \frac{n}{n-1} \frac 1 {(k-1)!} \le \frac 2 {(k-1)!}$ for $n \ge 2$ . Thus we get a suitable domination.

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Figaro

#4 h Apr 1, 2025, 5:47 AM • 2 Y

Y by Snoop76, Svyatoslav

$\lim_{n \to \infty} \sum_{k=1}^n \frac {n}{k(n-k)!}=\lim_{n \to \infty} \sum_{k=0}^{n-1} \frac {n}{(n-k)k!}=\lim_{n \to \infty} \sum_{k=0}^{n-1} \frac {n-k+k}{(n-k)k!}=$

$=\lim_{n \to \infty} \sum_{k=0}^{n-1} \frac {1}{k!}+\lim_{n \to \infty} \sum_{k=0}^{n-1} \frac {k}{(n-k)k!}=e+\lim_{n \to \infty} \sum_{k=1}^{n-1} \frac {1}{(n-k)(k-1)!}$

It remains to show that the limit we are left with is 0. For the sake of convenience we write

$\lim_{n \to \infty} \sum_{k=1}^{n-1} \frac {1}{(n-k)(k-1)!}=\lim_{n \to \infty} \sum_{k=1}^{n^2-1} \frac {1}{(n^2-k)(k-1)!}=$

$=\lim_{n \to \infty} \sum_{k=1}^{n} \frac {1}{(n^2-k)(k-1)!}+\lim_{n \to \infty} \sum_{k=n+1}^{n^2-1} \frac {1}{(n^2-k)(k-1)!}$

Now we can squeeze:

$0 \leq \sum_{k=1}^{n} \frac {1}{(n^2-k)(k-1)!} \leq \sum_{k=1}^{n} \frac {1}{n^2-n} = \frac{1}{n-1} \to 0$

and

$0 \leq \sum_{k=n+1}^{n^2-1} \frac {1}{(n^2-k)(k-1)!} \leq \sum_{k=n+1}^{n^2-1} \frac {1}{n!} = \frac{n^2-n-1}{n!} \leq \frac{1}{(n-2)!} \to 0$

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#5 h Apr 1, 2025, 4:39 PM • 2 Y

Y by MS_asdfgzxcvb, Snoop76

$\sum_{k=1}^n \frac n {k(n-k)!}\sim e\,+\frac e{n-1}\sum_{k=0}^\infty\left(\frac{\partial^k}{\partial\alpha^k}\,\bigg|_{\alpha=0}e^{e^\alpha}\right)\frac1{(n-1)^k}=e\,\left(1+\frac 1{n-1}+\frac 1{(n-1)^2}+\frac 2{(n-1)^3}+\frac5{(n-1)^4}+\frac {15}{(n-1)^5}+...\right)$

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#6 h Apr 3, 2025, 4:05 AM

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Figaro wrote:

$\lim_{n \to \infty} \sum_{k=1}^n \frac {n}{k(n-k)!}=\lim_{n \to \infty} \sum_{k=0}^{n-1} \frac {n}{(n-k)k!}=\lim_{n \to \infty} \sum_{k=0}^{n-1} \frac {n-k+k}{(n-k)k!}=$

$=\lim_{n \to \infty} \sum_{k=0}^{n-1} \frac {1}{k!}+\lim_{n \to \infty} \sum_{k=0}^{n-1} \frac {k}{(n-k)k!}=e+\lim_{n \to \infty} \sum_{k=1}^{n-1} \frac {1}{(n-k)(k-1)!}$

It remains to show that the limit we are left with is 0. For the sake of convenience we write

$\lim_{n \to \infty} \sum_{k=1}^{n-1} \frac {1}{(n-k)(k-1)!}=\lim_{n \to \infty} \sum_{k=1}^{n^2-1} \frac {1}{(n^2-k)(k-1)!}=$

$=\lim_{n \to \infty} \sum_{k=1}^{n} \frac {1}{(n^2-k)(k-1)!}+\lim_{n \to \infty} \sum_{k=n+1}^{n^2-1} \frac {1}{(n^2-k)(k-1)!}$

Now we can squeeze:

$0 \leq \sum_{k=1}^{n} \frac {1}{(n^2-k)(k-1)!} \leq \sum_{k=1}^{n} \frac {1}{n^2-n} = \frac{1}{n-1} \to 0$

and

$0 \leq \sum_{k=n+1}^{n^2-1} \frac {1}{(n^2-k)(k-1)!} \leq \sum_{k=n+1}^{n^2-1} \frac {1}{n!} = \frac{n^2-n-1}{n!} \leq \frac{1}{(n-2)!} \to 0$

For your limit $\lim_{n \to \infty} \sum_{k=1}^{n-1} \frac {1}{(n-k)(k-1)!},$ you didn't prove the exisence of this limit. If the limit exists, you method is good. Actually, you can prove the exisence by $\epsilon-N$ method.

Or: Note that $\sum_{k=1}^{n-1} \frac {1}{(n-k)(k-1)!}=\frac1{n-1}+\frac1{n-2}+\sum_{k=3}^{n-1} \frac {1}{(n-k)(k-1)!}\leq\frac1{n-1}+\frac1{n-2}+\sum_{k=3}^{n-1} \frac {1}{(n-k)(k-1)(k-2)},$ and
$\frac {1}{(n-k)(k-1)(k-2)}=\frac1{(n-k)(k-2)}-\frac1{(n-k)(k-1)} =\frac1{n-2}\left(\frac1{n-k}+\frac1{k-2}\right)-\frac1{n-1}\left(\frac1{n-k}+\frac1{k-1}\right),$ which imply that $\lim_{n \to \infty} \sum_{k=1}^{n-1} \frac {1}{(n-k)(k-1)!}=0.$

This post has been edited 1 time. Last edited by Alphaamss, Apr 3, 2025, 5:22 AM
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Figaro

#7 h Apr 3, 2025, 12:37 PM

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Thank you for your contribution, Alphaamss. Maybe I didn't make myself clear. I'll try to show in more detail what I did in #4.

(1) I want to show that $\lim_{n \to \infty} \sum_{k=1}^{n-1} \frac {1}{(n-k)(k-1)!}=0$ . I do not assume that the limit exists.

(2) Purely for the sake of convenience I'll have $n^2$ instead of $n$ , which obviously does not alter the limit (should it exist) and get
$\lim_{n \to \infty} \sum_{k=1}^{n^2-1} \frac {1}{(n^2-k)(k-1)!}$
You can leave out this step, but then you have to deal with square roots and the floor function. Not a problem, but maybe not so convenient.

(3) I take the FINITE sum $\sum_{k=1}^{n^2-1} \frac {1}{(n^2-k)(k-1)!}$ and split it into two FINITE sums $\sum_{k=1}^{n} \frac {1}{(n^2-k)(k-1)!}$ and $\sum_{k=n+1}^{n^2-1} \frac {1}{(n^2-k)(k-1)!}$ .

(4) I establish the two inequalities $0 \leq \sum_{k=1}^{n} \frac {1}{(n^2-k)(k-1)!} \leq \frac{1}{n-1}$ and $0 \leq \sum_{k=n+1}^{n^2-1} \frac {1}{(n^2-k)(k-1)!} \leq \frac{1}{(n-2)!}$ .

(5) We know that $\lim_{n \to \infty} 0$ exists and is 0 and $\lim_{n \to \infty} \frac{1}{n-1}$ exists and is 0. Using the squeeze theorem, I conclude that $\lim_{n \to \infty} \sum_{k=1}^{n} \frac {1}{(n^2-k)(k-1)!}$ EXISTS and is 0. Same for the other sum.

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#8 h Apr 4, 2025, 2:41 AM

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Figaro wrote:

Thank you for your contribution, Alphaamss. Maybe I didn't make myself clear. I'll try to show in more detail what I did in #4.

(1) I want to show that $\lim_{n \to \infty} \sum_{k=1}^{n-1} \frac {1}{(n-k)(k-1)!}=0$ . I do not assume that the limit exists.

(2) Purely for the sake of convenience I'll have $n^2$ instead of $n$ , which obviously does not alter the limit (should it exist) and get
$\lim_{n \to \infty} \sum_{k=1}^{n^2-1} \frac {1}{(n^2-k)(k-1)!}$
You can leave out this step, but then you have to deal with square roots and the floor function. Not a problem, but maybe not so convenient.

(3) I take the FINITE sum $\sum_{k=1}^{n^2-1} \frac {1}{(n^2-k)(k-1)!}$ and split it into two FINITE sums $\sum_{k=1}^{n} \frac {1}{(n^2-k)(k-1)!}$ and $\sum_{k=n+1}^{n^2-1} \frac {1}{(n^2-k)(k-1)!}$ .

(4) I establish the two inequalities $0 \leq \sum_{k=1}^{n} \frac {1}{(n^2-k)(k-1)!} \leq \frac{1}{n-1}$ and $0 \leq \sum_{k=n+1}^{n^2-1} \frac {1}{(n^2-k)(k-1)!} \leq \frac{1}{(n-2)!}$ .

(5) We know that $\lim_{n \to \infty} 0$ exists and is 0 and $\lim_{n \to \infty} \frac{1}{n-1}$ exists and is 0. Using the squeeze theorem, I conclude that $\lim_{n \to \infty} \sum_{k=1}^{n} \frac {1}{(n^2-k)(k-1)!}$ EXISTS and is 0. Same for the other sum.

Let $a_n=\sum_{k=1}^{n-1} \frac {1}{(n-k)(k-1)!}$ , you dealed with $\{a_{n^2}\}$ which is a subsequence of $\{a_n\}$ .
But $\{a_{n^2}\}$ is convergent doesn't necessarily imply $\{a_n\}$ is convergent, unless there are some extra properties of $\{a_n\}$ known (for example, $\{a_n\}$ is increasing).

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