2
$\begingroup$

Let $N>0$ be a large integer, and $n<N$, then how to simply the following sum $$\sum\limits_{k=1}^n\frac{N-n+k}{(N-k+1)(N-k+1)(N-k)}.$$ Thank you very much, guys.

Actually for another similar sum $\sum\limits_{k=1}^n\frac{1}{(N-k+1)(N-k)}=\sum\limits_{k=1}^n\frac{1}{N-k}-\frac{1}{N-k+1}=\frac{1}{N-n}-\frac{1}{N}$, I know the trick. But adding one term of such thing, $\frac{N-n+k}{N-k+1}$, it becomes difficult.

So, thanks a million for any clue.

$\endgroup$
1
  • $\begingroup$ Do you mean the repeated factor in the denominator? $\endgroup$ Oct 5 '16 at 3:14
1
$\begingroup$

I am not sure that you will like it.

$$S_n=\sum\limits_{k=1}^n\frac{N-n+k}{(N-k+1)^2(N-k)}$$ $$S_n=\frac{n (n-2 N)}{N (n-N)}+(n-2 N-1)\, \big(\psi ^{(1)}(-N)-\psi ^{(1)}(n-N)\big)$$ where appears the first derivative of the digamma function. I do not think that this could be further simplified. The trouble is that $\psi ^{(1)}(m)$ is undefined for $m\leq 0$.

May be, you could prefer the following. Considering for large values of $N$ $$\frac{N-n+k}{(N-k+1)^2(N-k)}=\left(\frac{1}{N}\right)^2+\frac{4 k-n-2}{N^3}+\frac{9 k^2-3 k n-10 k+2 n+3}{N^4}+\frac{16 k^3-6 k^2 n-28 k^2+8 k n+18 k-3 n-4}{N^5}+\frac{25 k^4-10 k^3 n-60 k^3+20 k^2 n+60 k^2-15 k n-28 k+4 n+5}{N^6}+O\left(\frac{1}{N^7}\right)$$ and now summing from $k=1$ to $k=n$, we should get, as an approximation, $$S_n=\frac{n \left(N \left(6 N^3-3 N+2\right)+1\right)}{6 N^6}+\frac{n^2 \left(6 N^3-6 N+5\right)}{6 N^6}+\frac{n^3 (N (9 N-2)-10)}{6 N^6}+\frac{n^4 (12 N-5)}{6 N^6}+\frac{5 n^5}{2 N^6}$$

For sure, we could add more terms for higher accuracy. For illustration purposes, I used $N=1000$ and varied $n$. The following table reports the decimal values of the exact sum and of the ugly approximation. $$\left( \begin{array}{ccc} n & \text{exact} & \text{approximation} \\ 50 & 0.00005270 & 0.00005270 \\ 100 & 0.00011173 & 0.00011173 \\ 150 & 0.00017880 & 0.00017876 \\ 200 & 0.00025625 & 0.00025600 \\ 250 & 0.00034721 & 0.00034618 \\ 300 & 0.00045610 & 0.00045276 \\ 350 & 0.00058916 & 0.00057993 \\ 400 & 0.00075548 & 0.00073276 \\ 450 & 0.00096866 & 0.00091727 \\ 500 & 0.00124975 & 0.00114053 \end{array} \right)$$

$\endgroup$
5
  • $\begingroup$ this doesn't answer the question ; the title specifies "simply" and not the brute calculus ... $\endgroup$
    – user354674
    Oct 5 '16 at 4:35
  • $\begingroup$ @igael. There is no simple expression. The exact result is the first one I gave. $\endgroup$ Oct 5 '16 at 4:37
  • $\begingroup$ if there is no simple expression, the answer is simply "no" $\endgroup$
    – user354674
    Oct 5 '16 at 4:46
  • $\begingroup$ og N ? is it a log ? yes, perhaps there is no trick ... $\endgroup$
    – user354674
    Oct 5 '16 at 5:12
  • 1
    $\begingroup$ @igael. Sorry for the typo og was supposed to be of. On French keyboards, f and g are next to eachother. $\endgroup$ Oct 5 '16 at 5:37
0
$\begingroup$

With the same method which you have used above you get

$$\sum\limits_{k=1}^n \frac{N-n+k}{(N-k+1)^2(N-k)}=\frac{n(2N-n)}{N(N-n)}-(2N+1-n)\sum\limits_{k=1}^n \frac{1}{( N-k+1)^2}$$

Hints:

$\enspace N-n+k=(2N+1-n)-(N-k+1)$

$\enspace \displaystyle \frac{1}{(N-k+1)^2(N-k)}=\frac{1}{N-k}-\frac{1}{N-k+1}-\frac{1}{(N-k+1)^2} $

The closed form for $\sum\limits_{k=1}^n \frac{1}{( N-k+1)^2}$ is:

$$\sum\limits_{k=1}^n \frac{1}{( N-k+1)^2}= \sum\limits_{k=1}^N \frac{1}{k^2} -\sum\limits_{k=1}^{N-n} \frac{1}{k^2}=$$$$((\frac{1}{N!} \begin{bmatrix} N+1 \\ 2 \end{bmatrix})^2-\frac{2}{N!} \begin{bmatrix} N+1 \\3 \end{bmatrix})-((\frac{1}{(N-n)!} \begin{bmatrix} N-n+1 \\ 2 \end{bmatrix})^2-\frac{2}{(N-n)!} \begin{bmatrix} N-n+1 \\3 \end{bmatrix}) $$

where $\begin{bmatrix} n \\ k \end{bmatrix}$ is called unsigned Stirling number of the first kind .

$\endgroup$
0
$\begingroup$

$\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\color{#f00}{% \sum_{k = 1}^{n}{N - n + k \over \pars{N - k + 1}\pars{N - k + 1}\pars{N - k}}} \\[5mm] = &\ \pars{2N - n}\pars{\sum_{k = 1}^{n}{1 \over k - N - 1} - \sum_{k = 1}^{n}{1 \over k - N}} - \pars{2N - n + 1}\sum_{k = 1}^{n}{1 \over \pars{k - N - 1}^{2}} \\[5mm] = &\ {\pars{2N - n}n \over N\pars{N - n}} - \pars{2N - n + 1}\sum_{k = 1}^{n}{1 \over \pars{k - N - 1}^{2}} \end{align}

Note that

$$\left\{\begin{array}{rcl} \ds{\sum_{k = 1}^{n}{1 \over k + a}} & \ds{=} & \ds{\sum_{k = 0}^{n - 1}{1 \over k + 1 + a} = \sum_{k = 0}^{\infty}\pars{{1 \over k + 1 + a} - {1 \over k + n + 1 + a}} = H_{n + a} - H_{a}} \\[2mm] \ds{\sum_{k = 1}^{n}{1 \over \pars{k + a}^{2}}} & \ds{=} & \ds{\partiald{}{a}\pars{H_{a} - H_{n + a}} = \Psi\,'\pars{1 + a} - \Psi\,'\pars{1 + n + a}} \end{array}\right. $$

$\ds{H_{z}}$ is the Harmonic Number and $\ds{\Psi\,'}$ is the Trigamma Function.


Then, \begin{align} &\color{#f00}{% \sum_{k = 1}^{n}{N - n + k \over \pars{N - k + 1}\pars{N - k + 1}\pars{N - k}}} \\[5mm] = &\ \color{#f00}{{\pars{2N - n}n \over N\pars{N - n}} - \pars{2N - n + 1}\bracks{\Psi\,'\pars{-N} - \Psi\,'\pars{-N - n}}} \end{align}

Any issue with the Trigamma's argument signs can be deal with the Euler Reflection Formula or/and the Recurrence Formula:

$$ \left\{\begin{array}{rcl} \ds{\Psi\,'\pars{z}} & \ds{=} & \ds{-\Psi\,'\pars{1 - z} + \pi^{2}\csc^{2}\pars{\pi z}} \\[2mm] \ds{\Psi\,'\pars{z + 1}} & \ds{=} & \ds{\Psi\,'\pars{z} - {1 \over z^{2}}} \end{array}\right. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.