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I am stuck summing a simple infinity series

$$\sum_{n=0}^{\infty}\frac{x^{n+3}}{(n+3)!}$$

I know that

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

And I presume I should divide my expresion into some kind partial fractions right? Something like

$$\sum_{n=0}^{\infty}\frac{x^{n+3}}{(n+3)!}=\frac{ax^n}{n!}+\frac{bx}{(n+1)}+\frac{cx}{(n+2)}+\frac{dx}{(n+3)}$$

But I can't find a wat of how to solve it for $a,b,c,d$. Possibly, another - simpler way exsits. Any hints? Thanks!

Sarunas
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  • $$ f(x) = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + \sum_{n=1}^{\infty} \frac{x^n}{n!} $$

    $$ \Rightarrow f'(x) = \sum_{n=1}^{\infty} \frac{x^{n-1}}{(n-1)!} = f(x) \Rightarrow \frac{f'(x)}{f(x)} = 1 $$

    $$ \Rightarrow \int \frac{df}{f} = \int dx = x + c \Rightarrow \ln f(x) = x + c \Rightarrow f(x) = c_1e^x $$

    $$ but : f(0) = 1 \Rightarrow f(x) = e^x $$

     – what'sup Aug 13 '13 at 15:12

3 Answers3

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You have $$\begin{align} \sum_{n=0}^{\infty}\frac{x^{n+3}}{(n+3)!} &= \sum_{n = 3}^{\infty} \frac{x^n}{n!}\\ &= \sum_{n = 0}^{\infty} \frac{x^n}{n!} -\frac{x^0}{0!} - \frac{x^1}{1!} - \frac{x^2}{2!}. \end{align}$$

Thomas
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$$\sum_{n=0}^{\infty}\frac{x^{n+3}}{(n+3)!} = \sum_{n = 3}^{\infty} \frac{x^n}{n!}=\left(\sum_{n = 0}^{\infty} \frac{x^n}{n!}\right)-1-x-\frac{x^2}{2}=e^x-1-x-\frac{x^2}{2}$$.

Adi Dani
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Hint: Shift $n+3\to k$. So $\sum_0^{\infty}f(n)=\sum_3^{\infty}f(k)$ and so summation is $$\left(\sum_0^{\infty}f(k)+x^2/2!+x/1!+1\right)=\text{e}^x-(x^2/2!+x/1!+1)$$

Mikasa
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