Limit of $a(k)$ in $ \sum_{k=1}^n \frac{a_k}{(n+1-k)!} = 1 $

Question

For n = 1, 2, 3 ... (natural number)

$ \sum_{k=1}^n \frac{a_k}{(n+1-k)!} = 1 $

$ a_1 = 1, \ a_2 = \frac{1}{2}, \ a_3 = \frac{7}{12} \cdots $

What is the limit of {$ a_k $}

$ \lim_{k \to \infty} a_k $ = ?

I have no idea where to start.

Answer 1

Note that $$ \begin{align} \frac{x}{1-x} &=\sum_{n=1}^\infty x^n\\ &=\sum_{n=1}^\infty\sum_{k=1}^n\frac{a_k}{(n-k+1)!}x^n\\ &=\sum_{k=1}^\infty\sum_{n=k}^\infty\frac{a_k}{(n-k+1)!}x^n\\ &=\sum_{k=1}^\infty\sum_{n=0}^\infty\frac{a_k}{(n+1)!}x^{n+k}\\ &=\frac{e^x-1}{x}\sum_{k=1}^\infty a_kx^k\tag{1} \end{align} $$ Therefore, $$ \sum_{k=1}^\infty a_kx^k=\frac{x^2}{(e^x-1)(1-x)}\tag{2} $$ If $a_k$ limit to some $b$, then as $x\to1$ we would have $$ \begin{align} b &=\lim_{x\to1^-}\frac{\displaystyle\sum_{k=1}^\infty a_kx^k}{\displaystyle\sum_{n=0}^\infty x^k}\\ &=\lim_{x\to1^-}\frac{x^2}{e^x-1}\\[9pt] &=\frac1{e-1}\tag{3} \end{align} $$ Now that we have an idea of what the limit would be, let's try to prove it.

From the defining formula, $$ a_n=1-\sum_{k=1}^\infty\frac{a_{n-k}}{(k+1)!}\tag{4} $$ where we define $a_k=0$ for $k\le0$.

Thus, if we let $c_n=a_n-\frac1{e-1}$, then for $n\gt1$, we have $$ c_n=-\sum_{k=1}^\infty\frac{c_{n-k}}{(k+1)!}\tag{5} $$ Note that if $|c_k|\lt c\,(4/5)^k$ for $k\lt n$ then $$ \begin{align} |c_n| &\le\sum_{k=1}^\infty\frac{c\,(4/5)^{n-k}}{(k+1)!}\\ &=c\,(4/5)^n\sum_{k=1}^\infty\frac{(4/5)^{-k}}{(k+1)!}\\ &=c\,(4/5)^n\frac45\left(e^{5/4}-1-\frac54\right)\\[9pt] &\le c\,(4/5)^n\tag{6} \end{align} $$ Since $c_1=\frac{e-2}{e-1}$ and $c_n=-\frac1{e-1}$ for $n\le0$, we can use $c=\frac1{e-1}$ in $(6)$. Therefore, $$ \begin{align} \left|\,a_n-\frac1{e-1}\,\right| &=|c_n|\\ &\le\frac{(4/5)^n}{e-1}\tag{7} \end{align} $$ Therefore, $$ \lim_{n\to\infty}a_n=\frac1{e-1}\tag{8} $$

Answer 2

Put $\displaystyle f(x)=\sum_{k\geq 1} a_k x^{k-1}$, $\displaystyle g(x)=\sum_{l\geq 0} \frac{x^l}{(l+1)!}=\frac{\exp(x)-1}{x}$, and $\displaystyle h(x)=f(x)g(x)=\sum_{n\geq 0}c_n x^n$. The coefficient $c_n$ is equal to $1$ if $n=0$, and for $n\geq 1$: $$c_n=\sum_{k-1+l=n, k\geq 1, l\geq 0}\frac{a_k}{(l+1)!}=\sum_{k=1}^{n}\frac{a_k}{(n+1-k)!}=1$$

Hence:

$$(\frac{\exp(x)-1}{x})f(x)=\sum_{n\geq 0}x^n=\frac{1}{1-x}$$ and $\displaystyle f(x)=\frac{x}{(1-x)(\exp(x)-1)}$. Put now $\displaystyle r(x)=(1-x)f(x)=\sum_{m\geq 0}d_m x^m=\frac{x}{\exp(x)-1}$. We have $d_0=a_1=1$, and for $m\geq 1$, $d_m=a_{m+1}-a_{m}$.

The radius of convergence of the power series $r(x)$ is $\geq 2\pi$, because it is regular at $x=0$, and the function $\exp(z)-1$ is non zero for $z\in \mathbb{C}$, $z\not = 0$ and $|z|<2\pi$. Hence we can put $x=1$ in the formula $\displaystyle r(x)=\frac{x}{\exp(x)-1}$, we have hence $\displaystyle r(1)=\frac{1}{e-1}$.

Now $r(1)$ is the limit of $d_0+\cdots d_{n-1}=t_n$ if $n\to +\infty$; but one can easily see that $t_n=a_n$, so $\displaystyle a_n\to \frac{1}{e-1}$.