Closed form expression of $\frac{x^{n+1}}{(n+1)!}+\frac{x^{n+2}}{(n+2)!}+...=e^x-(1+x+...+\frac{x^n}{n!})$

Question

$1+x+\frac{x^2}{2!}+...\frac{x^n}{n!}$ it is clear that it is sum of the finite terms of exponential series.

$1+x+...+\frac{x^n}{n!}+....=e^x$

Can we write this as $\frac{x^{n+1}}{(n+1)!}+\frac{x^{n+2}}{(n+2)!}+...=e^x-(1+x+...+\frac{x^n}{n!})$.

Is there any explicit formula for this?

Answer 1

Yes. For any convergent series $\sum_{n=0}^\infty a_n$, if $s$ is its sum and $N\in\mathbb N$, we have$$\sum_{n=N+1}^\infty a_n=s-\sum_{n=0}^Na_n.$$

Answer 2

Weird.

First of all the exponential series, as well as every series-expanded function (provided its convergence), has an infinite amount of terms, indeed it's almost always something like

$$f(x) = \sum_{k = \{0,1\}}^{+\infty} c_k x^k$$

For some coefficient $c_k$.

In your case you just have

$$e^{x} = \sum_{k = 0}^{+\infty} \frac{x^k}{k!}$$

If you want to write it in your way, you can do this

$$\sum_{k = n+1}^{+\infty} \frac{x^k}{k!} = e^x - \sum_{k = 0}^{n}\frac{x^k}{k!}$$

The left hand term can be viewed in terms of special functions in terms of the Euler incomplete Gamma Function as follows:

$$\sum_{k = n+1}^{+\infty} \frac{x^k}{k!} = e^x \left(1-\frac{\Gamma (n+1,x)}{\Gamma (n+1)}\right)$$

More about Euler Gamma: https://en.wikipedia.org/wiki/Gamma_function

Incomplete Gamma: https://en.wikipedia.org/wiki/Incomplete_gamma_function

Answer 3

Say $p_n(x)=\sum_{k=0}^n\frac{x^k}{k!}$. Then $p_n$ is just the n-th Taylor polynomial for the exponential. So $e^x-p_n(x)$ is the "remainder"; various forms of Taylor's theorem give various expressions for the remainder. For example the "integral form": $$e^x-p_n(x)=\int_0^x\frac{e^t}{n!}(x-t)^n\,dt.$$

Answer 4

For what it is worth.

Consider the integral operator

$$I(f):=\int_0^x f(t)\,dt.$$ It is easy to show that your function is

$$g(x)=I^n(e^x)$$ where exponentiation denotes iteration.

Now using the Laplace transform,

$$g(x)=\mathcal L^{-1}\left(\frac1{s^n(s-1)}\right),$$

which can be written as

$$g(x)=\frac1{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}\frac{e^{sx}}{s^n(s-1)}ds$$ where $\gamma>1.$

Answer 5

The "explicit formula" for $\sum_{k=0}^n x^k/k!$ is $e^x \Gamma(n+1,x)/n!$, where $\Gamma(\ldots,\ldots)$ is the incomplete Gamma function.

Answer 6

Using the incomplete gamma function $$\sum_{i=1}^m\frac{x^{n+i}}{(n+i)!}=e^x \left(\frac{\Gamma (m+n+1,x)}{(m+n)!}-\frac{\Gamma (n+1,x)}{n!}\right)$$ $$\sum_{i=1}^\infty\frac{x^{n+i}}{(n+i)!}=e^x \left(1-\frac{\Gamma (n+1,x)}{n!}\right)$$