I have to find the sum of this series $$\sum_{n=1}^{+\infty}\frac{x^{n-1}}{n!}$$

Using integral, I got $$\int\sum_{n=1}^{+\infty}\frac{x^{n-1}}{n!}=\sum_{n=1}^{+\infty}\frac{x^{n}}{n\cdot n!}$$

I know that $$\sum_{n=1}^{+\infty}\frac{x^{n}}{n!}=e^{x}$$

But my problem is about this $n\cdot n!$. How can I solve it?


  • $\begingroup$ The answer is closely related to $e^x-1$. Write down the series for this, and the answer will pop out. The integration will turn out to be unnecessary. $\endgroup$ – André Nicolas Jun 15 '14 at 20:34
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    $\begingroup$ Try something simpler than integration (like adding 1 and/or multiplying by $x$). $\endgroup$ – Stephen Montgomery-Smith Jun 15 '14 at 20:34
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    $\begingroup$ By the way, your last sum is $e^x-1$, not $e^x$ (since you started the sum at $n=1$). Divide through by $x$ (and also state what your first sum is when $x=0$) ... $\endgroup$ – David Mitra Jun 15 '14 at 20:35
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    $\begingroup$ just multiply and divide the first function by x ! Indeed for $x\neq0$ : $$\sum_{n=1}^{+\infty}\frac{x^{n-1}}{n!}=\frac1x\sum_{n=1}^{+\infty}\frac{x^n}{n!}=\frac1x(e^x-1)$$ $\endgroup$ – Fardad Pouran Jun 15 '14 at 20:36
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    $\begingroup$ @mvfs314 Just as long as whatever you're multiplying or dividing doesn't have an $n$ in it. Then it's just basically the distributive property. $\endgroup$ – Mike Jun 15 '14 at 20:43

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