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Let for all $x>0$ $$ f(x)=\sum_{n=0}^{+\infty}\frac{1}{x(x+1)\dots(x+n)}$$

Can you prove that for all $x>0$ $$f(x)= e \sum_{n=0}^{+\infty}\frac{(-1)^n}{(x+n)n!} $$ this is a question in a test for undergraduate students.

I checked that the series that defines $f$ converges. Moreover i proved that it uniformly converges in every interval of the form $[a,+\infty[$ with $a>0$.

One of my attempts to solve the exercise was to trying to differentiate both series and see if the expression of the derivatives was easier to handle. but I didn't get anywhere. Any suggestions?

irbag
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2 Answers2

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Apply partial fraction expansion to the $n^{th}$ term of the series of $f(x)$: $$ \frac{1}{x \cdot (x+1) \cdots (x+n)} = \frac{A_0}{x} + \frac{A_1}{x+1} + \cdots + \frac{A_n}{x+n} $$

The coefficients are easy to find: $$ A_k = \frac{(-1)^k}{(n-k)! \cdot k!} $$

Rewrite the series for $f(x)$ using the partial fraction expansion, interchange the summation symbols and simplify:

\begin{align} f(x) &= \sum_{n=0}^{\infty} \sum_{k=0}^{n} \frac{(-1)^k}{(n-k)! \cdot k!} \frac{1}{x+k} \\ &= \sum_{k=0}^{\infty} \sum_{n=k}^{\infty} \frac{(-1)^k}{(n-k)! \cdot k!} \frac{1}{x+k} \\ &= \sum_{k=0}^{\infty}\frac{(-1)^k}{k!} \frac{1}{x+k} \sum_{n=k}^{\infty} \frac{1}{(n-k)!} \\ &= \sum_{k=0}^{\infty}\frac{(-1)^k}{k!} \frac{1}{x+k} \sum_{n=0}^{\infty} \frac{1}{n!} \\ &= e \sum_{k=0}^{\infty}\frac{(-1)^k}{k!} \frac{1}{x+k} \end{align}

Metin Ersin Arıcan
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Using partial fractions, we have \begin{multline} \sum_{n=0}^\infty \prod_{i=0}^n\frac{1}{x+i} = \sum_{n=0}^\infty\sum_{i=0}^n\frac{(-1)^i}{i!(n-i)!(x+i)} =\sum_{n=0}^\infty\sum_{i=n}^\infty\frac{(-1)^n}{n!(i-n)!(x+n)} \\ = \sum_{n=0}^\infty\frac{(-1)^n}{(x+n)n!}\sum_{i=n}^\infty \frac{1}{(i-n)!} = \sum_{n=0}^\infty\frac{(-1)^n}{(x+n)n!}\sum_{i=0}^\infty \frac{1}{i!} = e\sum_{n=0}^\infty\frac{(-1)^n}{(x+n)n!} \end{multline}

eyeballfrog
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