Is there any approximate form of $\sum_{n=N}^\infty \frac{x^n}{n!}$?

Question

Is there any approximate form of $\sum_{n=N}^\infty \frac{x^n}{n!}$ for any integer $N > 0$ and any real valued $x$? The series comes from a truncated Taylor expansion of $e^x$. I tried to change the term into a continuous term that becomes $$ \int_N^\infty \frac{x^n}{\Gamma(n + 1)}\ dn. $$ However, I can't find the analytical form (or approximate form) of the integral above.

Answer 1

Deriving the WolframAlpha result in Jjacquelin's comment:

Here is a probabilistic interpretation that relates the sum to an integral. For a Poisson process with rate $\lambda > 0$, the expression $e^{-\lambda} \sum_{n = N}^\infty \frac{\lambda^n}{n!}$ is the probability that there are $\ge N$ arrivals in the time interval $[0, 1]$. This is equivalent to the $N$th arrival time being $\le 1$. The $N$th arrival time follows the gamma distribution with shape $N$ and rate $\lambda$, and has density $f(t) = \frac{\lambda^N}{\Gamma(N)} t^{N-1} e^{-\lambda t}$. Thus we have $$e^{-\lambda} \sum_{n=N}^\infty \frac{\lambda^n}{n!} = \int_0^1 \frac{\lambda^N}{\Gamma(N)} t^{N-1} e^{-\lambda t} \, dt.$$ See here for a reference of the above equality, and note that one can also obtain this directly (without any probability stuff) by using integration by parts repeatedly. Doing integration by parts also proves the identity for non-positive $\lambda$.

As the Wikipedia article demonstrates, this integral can be expressed as $\gamma(N, \lambda)/\Gamma(N)$, where $\gamma$ is the lower incomplete gamma function. So, $$\sum_{n=N}^\infty \frac{\lambda^n}{n!} = e^{\lambda} \frac{\gamma(N, \lambda)}{\Gamma(N)}$$ and noting that $\Gamma(N) = \Gamma(N, \lambda) + \gamma(N, \lambda)$ draws the connection with the WolframAlpha result in the comments.

This is the best you can do for an exact expression (and arguably is not really much more amenable for analysis than the original sum). Depending on your needs, you could apply known bounds/asymptotics for the incomplete gamma functions if you wish.

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The Taylor series approach suggested by Marius S.L.:

Taylor's theorem implies that $$\sum_{n=N}^\infty \frac{x^n}{n!} = e^{\xi_N} \frac{x^N}{N!}$$ where $\xi_N$ is some number between $0$ and $x$. So if $x$ is fixed and known, a crude bound is $$\left|\sum_{n=N}^\infty \frac{x^n}{n!}\right| \le e^{\xi_n} \frac{|x|^N}{N!} \le \max\{1, e^x\} \frac{|x|^N}{N!}.$$

Answer 2

The sum you have is $$e^x-\sum_{n=0}^N\frac{x^n}{n!}$$And now we need to find the value of$$\sum_{n=0}^N\frac{x^n}{n!}$$ It doesn't seem possible to find this partial sum. But for large $N$ this gets closer to $e^x$, so a good approximation for your sum for large $N$ and small $x$ is $0$.

Another way to get this approximation is to see that the summand is getting smaller as $N$ is getting larger since the factorial is always bigger than any monomial that it divides.