Does someone know how to calculate sum $\sum_{k=1}^n \frac{1}{(n-k)!\cdot k} $? I was working something with matrices, i.e. calculating number of all cycles in Coates digraph, and got this weird expression(multiplied by $n!$) Thank you in advance.
$\begingroup$
$\endgroup$
4
-
$\begingroup$ Hm. With the multiplication by $n!$, each summand has the form $$\binom n k \cdot (k-1)!$$ I wonder if that means anything. $\endgroup$– PrincessEevCommented Mar 20, 2019 at 23:22
-
3$\begingroup$ Are you certain that it's $k$ in the denominator and not $k!$? (That's a factorial and a question mark, not an interrobang.) $\endgroup$– ArthurCommented Mar 20, 2019 at 23:23
-
$\begingroup$ Maybe try sticking an $x^k$ to it each term. Then it becomes some polynomial. If you differentiate this polynomial it takes on the form. $$ \frac{1}{(n-1)!} + \frac{1}{(n-2)!}x + ...$$ but it’s not clear if this direction is fruitful. $\endgroup$– Sidharth GhoshalCommented Mar 20, 2019 at 23:24
-
$\begingroup$ It looks like a convolution of $\frac{1}{k}$ with $\frac{1}{k!}$ - so its generating function would probably be something like $e^x \cdot (-\ln(1-x))$. $\endgroup$– Daniel ScheplerCommented Mar 20, 2019 at 23:33
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
I don't know of a closed form for your sequence, other than the hypergeometric
$$ {\frac {{\mbox{$_3$F$_1$}(1,1,1-n;\,2;\,-1)}}{ \left( n-1 \right) !}}$$
but the generating function is
$$ g(x) = \sum_{n=1}^\infty \sum_{k=1}^n \frac{x^n}{(n-k)! k} = -e^x \ln(1-x)$$
answered Mar 20, 2019 at 23:44