7

Context: The cube super root ssrt$_3(x)$ series expansion yielded part of it as:

$$\sum_{n=1}^\infty\frac{y^n}{n!}\sum_{m=0}^\infty\frac{(-n)^m}{m!}\sum_{k=0}^{n-1}\binom{n-1}k\left.\frac{d^kt^m}{dt^k}\right|_{t=0}\left.\frac{d^{n-1-k}e^{tm}}{d^{n-1-k}}\right|_{t=0}=\lim_{t\to0}\sum_{n=1}^\infty\frac{y^n}{n!}\sum_{m=0}^\infty \sum_{k=0}^{n-1}\frac{(-n)^m \Gamma(n) m^{n-k+1}t^{m-k}}{k!(m-k)!\Gamma(n-k)}$$

The $m=0$ term is ignored outside the sum. The inner sum uses Hypergeometric $U(a,b,x)$ and cancels $t^a$, so the limit was easier to find. Testing different cases of $m,n$ gave a pattern for the limit as a ratio of gamma functions:

$$\lim_{t\to0}\sum_{m,n=1}^\infty\frac{y^n m^nn^m}{m^{m+2}(m-1)!n!}U(-m,n-m,t)=\lim_{t\to0}\sum_{m,n=1}^\infty\frac{y^n m^nn^m(m-(n+t))!\sin(\pi(n+t))}{m^{m+2}\pi\Gamma(m)}=\sum_{m,n=1}^\infty\frac{y^n(-1)^mm^n n^{m-1}}{\Gamma(n-m)\Gamma(m)m^{m+2}}$$

Problem:

Also, switching the differentiation order, due to the general Leibniz rule, gives the an equal sum; part of it had this limit:

$$\sum_{n=1}^\infty\frac{y^n}{n!}\sum_{m=0}^\infty\frac{(-n)^m}{m!}\sum_{k=0}^{n-1}\binom{n-1}k\left.\frac{d^{n-1-k}t^m}{dt^{n-1-k}}\right|_{t=0}\left.\frac{d^ke^{tm}}{dt^k}\right|_{t=0}=\lim_{t\to0}\sum_{n=1}^\infty\sum_{m=0}^\infty\sum_{k=0}^{n-1}\frac{y^nm^k(-n)^m t^{k+m-n+1}}{(k+m-n+1)!\Gamma(n-k)k!n}$$

Taking $\lim\limits_{t\to0}t^{k+m-n+1}$ makes the sum diverge and seemingly removing any terms from the sum still makes the sum diverge, so we can’t substitute this confluent $\,_1\text F_1(a;b;x)$ identity:

$$\sum_{k=0}^{n-1}\frac{m^k(-n)^m t^{k+m-n+1}}{(k+m-n+1)!\Gamma(n-k)k!n}=\frac{(-n)^mt^{m-n+1}\,_1\text F_1(1-n;m-n+2;-mt) }{n!(m-n+1)!}$$

However, noticing that the truncated sums give a polynomial, the limit is the constant term of it and therefore Kronecker $\delta_x$ appears:

$$\lim_{t\to0}\sum\limits_{n\ge1}\sum\limits_{m\ge0}\sum\limits_{k=0}^{n-1}\frac{y^nm^k(-n)^mt^{k+m-n+1}}{(k+m-n+1)!\Gamma(n-k)k!n}= \sum\limits_{n\ge1}\sum\limits_{m\ge0}\sum\limits_{k=0}^{n-1}\frac{y^nm^k(-n)^m\delta_{k+m-n+1}}{(k+m-n+1)!\Gamma(n-k)k!n}$$

We can remove a sum like in

this post

or simplify, but how?

Тyma Gaidash
  • 12,081

1 Answers1

2

Quantile mechanics gives a method for evaluating Kronecker delta sums. The sum terms disappear unless $k=n-m-1$, so remove the $k$ sum and substitute:

$$\sum_{n=1}^\infty\sum\limits_{m=0}^\infty\sum\limits_{k=0}^{n-1}\frac{y^nm^k(-n)^m\delta_{k+m-n+1}}{(k+m-n+1)!\Gamma(n-k)k!n}=y+\sum_{n=1}^\infty\sum_{m=0}^\infty \frac{(-1)^my^n n^{m-1}}{m!\Gamma(n-m)m^{m-n+2}}$$

or the same but with $m=n-k-1$ gives:

$$\sum_{n=1}^\infty\sum\limits_{m=0}^\infty\sum\limits_{k=0}^{n-1}\frac{y^nm^k(-n)^m\delta_{k+m-n+1}}{(k+m-n+1)!\Gamma(n-k)k!n}=\sum_{n=1}^\infty\sum_{k=0}^{n-1}\frac{(-n)^{n-k-1}(n-k-1)^ky^n}{\Gamma(n-k)k!n}$$

Other derivations and solutions are welcome.

Тyma Gaidash
  • 12,081