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I have an infinite series such as $$ \sum_{k=1}^\infty \frac{k!}{(b+1)(b+2)...(b+k)}$$

I need to find all values of $b$ so this converges. So far I've found that cannot be negative, since that would cause a zero term in the denominator at some point. Also $b\ne 0$ since then the $k'th$ term would be $1$ and the series would diverge. If we assume that $b$ is a natural number, then we can write $$\sum_{k=1}^\infty \frac{k!}{\frac{1\cdot 2\cdot ...b(b+1)(b+2)...(b+k)}{1\cdot 2\cdot 3...\cdot b}} = \sum_{k=1}^\infty \frac{k!}{\frac{(k+b)!}{b!}} = \sum_{k=1}^\infty \frac{b!\cdot k!}{(k+b)!} = b!\sum_{k=1}^\infty \frac {1}{(k+1)(k+2)...(k+b)}$$ And that sum converges by comparison test when $b \gt 1$. But I have a feeling and WolframAlpha confirms that this converges for all numbers $b \gt 1$. How could I show that?

Markus Punnar
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2 Answers2

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If $b\leq 0$ then the main term does not converge to zero, so we may safely assume $b>0$.
With such assumption $$ \frac{k!}{(b+1)(b+2)\cdots(b+k)} = \frac{\Gamma(k+1)\Gamma(b+1)}{\Gamma(b+k+1)}=b\cdot B(b,k+1)=b\int_{0}^{1}x^{b-1}(1-x)^k\,dx $$ and $$ \sum_{k\geq 1}\frac{k!}{(b+1)(b+2)\cdots(b+k)}=b\int_{0}^{1}x^{b-2}(1-x)\,dx=\color{red}{\frac{1}{b-1}} $$ as soon as $b>1$. By Gautschi's inequality this is a necessary condition for convergence.

Jack D'Aurizio
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We have

$$\begin{align} a_k(b)&=\frac{k!}{(b+1)(b+2)\cdots (b+k)}\\\\ \end{align}$$

Applying Gauss's Test for series convergence reveals for $b>-1$

$$\frac{a_{k}(b)}{a_{k+1}(b)}=\frac{k+1+b}{k+1}=1+\frac{b}{k}+O\left(\frac1{k^2}\right)$$

whence we see that the series of interest converges when $b>1$ and diverges otherwise.

NOTE:

Gauss's Law can still be applied to the case $b<-1$, $-b\notin\mathbb{N}$, by factoring out the negative terms from the denominator and analyzing the series of the resulting positive summand.

Mark Viola
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