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For every value of $n_1$ i tested with, it came out as $0$. However I have no idea how to prove if it does or not for every value of ${n_1}$.

Note that ${n_1}$ has to be smaller than ${n_0}$ in the summation. $$\sum_{i=1}^{n_1} \frac{(-1)^{i+1}2^{{n_1}-i}i({n_0}({n_1}-2+i)+{n_1}(i-1)+i^2-i)}{i!({n_1}-i)!({n_0}+i)({n_0}+i-1)({n_0}+{n_1})}$$

For instance, plugging $2$ to ${n_1}$ and solving the summation gives $\frac{2{n_0}}{({n_0}+1){n_0}({n_0}+2)}-\frac{2(2{n_0}+4)}{2({n_0}+2)({n_0}+1)({n_0}+2)}$, which is equal to $0$.

I have done everything to my knowledge but am still stuck. Any tips?

Eduard
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  • I'm not quite sure about this, but if it is zero for every integer $n_1$, then I believe each term must be zero. But the dependence of the individual terms on $n_1$ itself makes me unsure. What values of $n_1$ have you tested it with, and what is the relation, if any, between $n_0$ and $n_1$? – The Count Oct 24 '18 at 02:35
  • I have tested with n1 equal to 1,2,3 and 4. I have coded an algorithm that gives the result of this summation given n0 and n1 and in every case the number is extremely close to zero but even for the values i've proven by hand that its zero, the algorithm still gives a number extremely close to zero, i suspect it is because im not working with the floating point numbers well – Eduard Oct 24 '18 at 03:52
  • $n_0$ shouldn't matter. it is just a constant. it should simplify in your algorithm without being defined. mathematica is probably a good program to test it in. – The Count Oct 24 '18 at 04:13
  • Well, I used the program you suggested, and it gave me the answer zero. However it didn't do any explaining. At least now I know it is zero and just have to prove it – Eduard Oct 24 '18 at 05:24
  • Well, it is certainly nice to know what to aim to prove, so that's a good thing. – The Count Oct 24 '18 at 19:51

1 Answers1

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Your sum equals $0$.

The sum can be simplified using the idea of telescoping series.

$$\begin{align}&\sum_{i=1}^{n_1} \frac{(-1)^{i+1}2^{{n_1}-i}i({n_0}({n_1}-2+i)+{n_1}(i-1)+i^2-i)}{i!({n_1}-i)!({n_0}+i)({n_0}+i-1)({n_0}+{n_1})} \\\\&=\frac{-2^{n_1}}{n_0+n_1}\sum_{i=1}^{n_1} \frac{(-\frac 12)^{i}({n_0}({n_1}-2+i)+{n_1}(i-1)+i^2-i)}{(i-1)!({n_1}-i)!({n_0}+i)({n_0}+i-1)}\tag1 \\\\&=\frac{-2^{n_1}}{n_0+n_1}\sum_{i=1}^{n_1}(f(i+1)-f(i)) \\\\&=\frac{-2^{n_1}}{n_0+n_1}(f(n_1+1)-f(1)) \\\\&=0\end{align}$$ where $$f(i)=\frac{(-\frac 12)^{i-1}(i-1)(n_1+1-i)}{(i-1)!(n_1+1-i)!(n_0+i-1)}$$

Added :

how does one find f(i)?

The denominator of the sum in $(1)$ is $(i-1)!({n_1}-i)!({n_0}+i)({n_0}+i-1)$.

To have this denominator in $f(i+1)-f(i)$, one can think that the denominator of $f(i)$ is $(i-2)!(n_1-i)!(n_0+i-1)$. This is because the denominator of $f(i+1)$ is $(i-1)!(n_1-i-1)!(n_0+i)$, so the denominator of $f(i+1)-f(i)$ can be $(i-1)!({n_1}-i)!({n_0}+i)({n_0}+i-1)$.

Having $(-\frac 12)^i$, one can think that $f(i)$ is of the form $$\frac{(-\frac 12)^{i-1}g(i)}{(i-2)!(n_1-i)!(n_0+i-1)}$$ where $g(i)$ is a polynomial on $i$.

Finally, comparing the terms gives $g(i)=1$ fortunately, one has $$f(i)=\frac{(-\frac 12)^{i-1}}{(i-2)!(n_1-i)!(n_0+i-1)}=\frac{(-\frac 12)^{i-1}(i-1)(n_1+1-i)}{(i-1)!(n_1+1-i)!(n_0+i-1)}$$

mathlove
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