Here is something that arose while playing around with a class of particular integrals.
Let $\displaystyle a_n= \int_0^1 \left( e^x - 1 - x - \frac{x^2}{2} - \cdots - \frac{x^n}{n!} \right) \, {\rm d}x$. Evaluate the series
$$\mathcal{S}=\sum_{n=0}^{\infty} a_n$$
Solution
Let us start with the key identity
$$ \int_{0}^{x} e^{-t} t^n \, {\rm d}t = n! e^{-x} \left ( e^x - 1 - x -\frac{x^2}{2} - \cdots - \frac{x^n}{n!} \right ) \tag{1} \label{*}$$
Integrating \eqref{*} we have that:
\begin{align*}
\int_{0}^{1}\left ( e^x - 1 - x - \frac{x^2}{2} - \cdots - \frac{x^n}{n!} \right ) \, {\rm d}x &= \frac{1}{n!} \int_{0}^{1} e^x \left (\int_{0}^{x} e^{-t} t^n \, {\rm d}t \right ) \, {\rm d}x \\
&= \frac{1}{n!} \left [ e^x \int_{0}^{x} e^{-t} t^n \, {\rm d}t \right ]_0^1 - \frac{1}{n!} \int_{0}^{1} x^n \, {\rm d}x\\
&= \frac{e}{n!} \int_{0}^{1} e^{-t} t^n \, {\rm d}t - \frac{1}{n! \left ( n+1 \right )}
\end{align*}
Thus for the sum we have successively:
\begin{align*}
\mathcal{S} &=\sum_{n=0}^{\infty} a_n \\
&= \sum_{n=0}^{\infty} \left [ \frac{e}{n!} \int_{0}^{1} e^{-t} t^n \, {\rm d}t - \frac{1}{n!(n+1)} \right ]\\
&= e \sum_{n=0}^{\infty} \frac{1}{n!} \int_{0}^{1}e^{-t} t^n \, {\rm d}t - \sum_{n=0}^{\infty} \frac{1}{n! (n+1)}\\
&=e \int_{0}^{1} e^{-t} \sum_{n=0}^{\infty} \frac{t^n}{n!} \, {\rm d}t - e + 1 \\
&= e \int_{0}^{1}e^{-t} e^t \, {\rm d}t + 1-e \\
&=1
\end{align*}
Some comments:
Let $\displaystyle a_n= \int_0^1 \left( e^x - 1 - x - \frac{x^2}{2} - \cdots - \frac{x^n}{n!} \right) \, {\rm d}x$. Evaluate the series
$$\mathcal{S}=\sum_{n=0}^{\infty} a_n$$
Solution
Let us start with the key identity
$$ \int_{0}^{x} e^{-t} t^n \, {\rm d}t = n! e^{-x} \left ( e^x - 1 - x -\frac{x^2}{2} - \cdots - \frac{x^n}{n!} \right ) \tag{1} \label{*}$$
Integrating \eqref{*} we have that:
\begin{align*}
\int_{0}^{1}\left ( e^x - 1 - x - \frac{x^2}{2} - \cdots - \frac{x^n}{n!} \right ) \, {\rm d}x &= \frac{1}{n!} \int_{0}^{1} e^x \left (\int_{0}^{x} e^{-t} t^n \, {\rm d}t \right ) \, {\rm d}x \\
&= \frac{1}{n!} \left [ e^x \int_{0}^{x} e^{-t} t^n \, {\rm d}t \right ]_0^1 - \frac{1}{n!} \int_{0}^{1} x^n \, {\rm d}x\\
&= \frac{e}{n!} \int_{0}^{1} e^{-t} t^n \, {\rm d}t - \frac{1}{n! \left ( n+1 \right )}
\end{align*}
Thus for the sum we have successively:
\begin{align*}
\mathcal{S} &=\sum_{n=0}^{\infty} a_n \\
&= \sum_{n=0}^{\infty} \left [ \frac{e}{n!} \int_{0}^{1} e^{-t} t^n \, {\rm d}t - \frac{1}{n!(n+1)} \right ]\\
&= e \sum_{n=0}^{\infty} \frac{1}{n!} \int_{0}^{1}e^{-t} t^n \, {\rm d}t - \sum_{n=0}^{\infty} \frac{1}{n! (n+1)}\\
&=e \int_{0}^{1} e^{-t} \sum_{n=0}^{\infty} \frac{t^n}{n!} \, {\rm d}t - e + 1 \\
&= e \int_{0}^{1}e^{-t} e^t \, {\rm d}t + 1-e \\
&=1
\end{align*}
Some comments:
- Maybe it worth looking at the more general cases:
$$\mathcal{S} = \sum_{n=0}^{\infty}\frac{a_n}{\left ( n! \right )^s} , \; \mathbb{N} \ni s \geq 0$$
These sums are more likely to involve the Bessel functions. - How about working with sums of the form:
$$\mathcal{S} = \sum_{n=0}^{\infty} \frac{1}{n!\left ( n+1 \right )(n+2)\cdots(n+k)} , \; \mathbb{N} \ni k \geq 1$$
Alright,
ReplyDeletelet us answer question $2$. Here we go:
\begin{align*}
\sum_{n=0}^{\infty} \frac{1}{n! (n+1)(n+2)\cdots(n+k)} &=\sum_{n=0}^{\infty} \frac{1}{n!}\left ( \prod_{m=1}^{k}\left ( n+m \right ) \right )^{-1} \\
&= \sum_{n=0}^{\infty}\frac{\Gamma(n+1)}{n! \Gamma (k+n+1)} \\
&=\sum_{n=0}^{\infty} \frac{1}{\Gamma(k+n+1)} \\
&=\sum_{n=0}^{\infty} \frac{1}{(k+n)!} \\
&\!\!\!\!\!\overset{k+n=m}{=\! =\! =\! =\!} \sum_{m=k}^{\infty} \frac{1}{m!} \\
&= \sum_{m=0}^{\infty} \frac{1}{m!} - \sum_{m=0}^{k-1} \frac{1}{m!} \\
&= e - \sum_{m=0}^{k-1} \frac{1}{m!}
\end{align*}
Of course the last finite sum relates to the upper incomplete Gamma function . One interesting property of this function is the following:
$$\Gamma(s, x) = (s-1)! e^{-x} \sum_{k=0}^{s-1} \frac{x^k}{k!} , \; \mathbb{N} \ni s \geq 1$$
Thus $\displaystyle \Gamma(s, 1) = (s-1)!e^{-1} \sum_{k=0}^{s-1} \frac{1}{k!}$.
As a bonus exercise we also get that:
ReplyDelete$$\sum_{n=0}^{\infty} \left(e - 1 -\sum_{k=0}^{n} \frac{1}{k! (k+1)} \right) =1 $$
since
\begin{align*}
a_n &= \int_{0}^{1} \left ( e^x - 1 - x - \frac{x^2}{2} - \cdots - \frac{x^n}{n!} \right ) \, {\rm d}x \\
&= \int_{0}^{1} \left ( e^x - \sum_{k=0}^{n} \frac{x^k}{k!} \right ) \, {\rm d}x\\
&= e - 1 - \sum_{k=0}^{n} \frac{1}{k!}\int_{0}^{1} x^k \, {\rm d}x \\
&=e - 1 - \sum_{k=0}^{n} \frac{1}{k! (k+1)}
\end{align*}
Now this new form actually tells is a more direct way that $\lim a_n =0$. The other form also revealed that fact to us but in a more indirect manner.
Using standard technics of generating functions we can even prove that:
Delete$$\sum_{n=0}^{\infty} \left ( \sum_{k=0}^{n} \frac{1}{k! (k+1)} \right ) x^n = \frac{e^x-1}{x \left ( 1-x \right )}$$
since $\displaystyle \sum_{k=0}^{\infty} \frac{x^k}{k! (k+1)} = \frac{e^x-1}{x}$.
The interval of convergence as well as the radius is pending.
Let us increase the difficult of the sum at comment #1. Let us see the sum
ReplyDelete$$\mathcal{S} = \sum_{n=0}^{\infty} \frac{1}{\left ( n! \right )^2 \prod \limits_{m=1}^{k} \left ( n+m \right )}$$
This indeed relates to the Bessel functions. Successively we have:
\begin{align*}
\mathcal{S} &=\sum_{n=0}^{\infty} \frac{1}{\left ( n! \right )^2 \prod \limits_{m=1}^{k} \left ( n+m \right )} \\
&= \sum_{n=0}^{\infty} \frac{1}{\left ( n! \right )^2} \left ( \prod_{m=1}^{k} (n+m) \right )^{-1}\\
&=\sum_{n=0}^{\infty} \frac{\Gamma(n+1)}{\left ( n! \right )^2 \Gamma(k+n+1)} \\
&=\sum_{n=0}^{\infty} \frac{1}{n! (n+k)!} \\
&=\mathfrak{mB}_k(2)
\end{align*}
where $\mathfrak{mB}$ is the modified Bessel function of first kind .