Does the series converge or diverge?

Question

I want to check, whether $$\sum\limits_{n=0}^{\infty }{\frac{n!}{(a+1)(a+2)...(a+n)}}$$

converges or diverges.

$a$ is a constant number

Ratio test

$$\begin{align} & \frac{a_{n}}{a_{n-1}}=\frac{n!}{(a+1)(a+2)...(a+(n-1))(a+n)}\cdot \frac{(a+1)(a+2)...(a+(n-1))}{(n-1)!}=\frac{n}{a+n} \\ & \underset{n\to \infty }{\mathop{\lim }}\,\frac{a_{n}}{a_{n-1}}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{n}{a+n}=1 \\ \end{align}$$

We know nothing

Root test

$$\sqrt[n]{a_{n}}=\sqrt[n]{\frac{n!}{(a+1)(a+2)...(a+(n-1))(a+n)}}=$$ I can't :(

I can only resolve the serie if "a" is a integer number

$$\begin{align} & \text{if }a\in \mathbb{Z} \\ & (a+1)(a+2)...(a+(n-1))(a+n)=(n+a)(a+(n-1))...(a+2)(a+1)= \\ & (n+a)(n+a-1)(n+a-2)...(a+2)(a+1)=\frac{(n+a)!}{a!} \\ & \Rightarrow a_{n}=\frac{a!n!}{(n+a)!} \\ & \text{if }a<0\Rightarrow \text{(}n+a)<n,a\in \mathbb{Z} \\ & \underset{n\to \infty }{\mathop{\lim }}\,a_{n}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{a!n!}{(n+a)!}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{a!n(n-1)...(n+a)!}{(n+a)!}= \\ & \underset{n\to \infty }{\mathop{\lim }}\,a!n(n-1)...(n+a-1)=\infty \\ & \Rightarrow \sum\limits_{n=0}^{\infty }{a_{n}=\infty } \\ & \text{if }a>1,a\in \mathbb{Z} \\ & a_{n}=\frac{a!n!}{(n+a)!}=\frac{a!n!}{(n+a)(n-1+a)...\underbrace{(n-a+a)!}_{n!}}=\frac{a!}{(n+a)(n-1+a)...(n+1)} \\ & \Rightarrow \\ & \frac{a!}{\underbrace{(n+a)(n-1+a)...(n+1)}_{a\text{ times}}}<\frac{a!}{(n+1)^{a}}\Rightarrow \\ & \sum\limits_{n=0}^{\infty }{\frac{a!}{(n+a)(n-1+a)...(n+1)}<\sum\limits_{n=0}^{\infty }{\underbrace{\frac{a!}{(n+1)^{a}}<\infty }_{a>1}}} \\ & \text{if }a=1 \\ & \sum\limits_{n=0}^{\infty }{\frac{a!n!}{(n+a)!}}=\sum\limits_{n=0}^{\infty }{\frac{a!n!}{(n+1)!}}=\sum\limits_{n=0}^{\infty }{\frac{a!}{n+1}}=\infty \\ \end{align}$$

But how i can resolve if "a" is not a integer number?

Answer 1

Since: $$(a+1)(a+2)\cdot\ldots\cdot(a+n)=\frac{\Gamma(a+n+1)}{\Gamma(a+1)}$$ assuming $\Re(a)>1$ we can write the original series as: $$ S = \sum_{n\geq 1}\frac{\Gamma(n+1)\Gamma(a+1)}{\Gamma(a+n+1)}=\sum_{n\geq 1}n\cdot B(n,a+1)=\sum_{n\geq 1}n\int_{0}^{1}x^{n-1}(1-x)^a\,dx \tag{1}$$ but $\sum_{n\geq 1}n x^{n-1}=\frac{1}{(1-x)^2}$ gives: $$ S = \int_{0}^{1}(1-x)^{a-2}\,dx = \int_{0}^{1}x^{a-2}\,dx = \frac{1}{a-1}.$$ In order to prove that the condition $\Re(a)>1$ is necessary for the convergence of the series, just notice that the Euler product for the $\Gamma$ function gives: $$\frac{\Gamma(n+1)}{\Gamma(n+a+1)}=\Theta\left(\frac{1}{n^a}\right)$$ hence the criterion for the convergence of the generalized harmonic series applies.

Answer 2

I don't know much about Gamma Function and summation under the integral so here is a more elementary proof.

• If $a>0$ we have $\ln a_n=-(\ln (1+\frac{a}{1})+\ln (1+\frac{a}{2})+\dots+\ln (1+\frac{a}{n}))$ and using

$$x\ge\ln(1+x)\ge x-\frac{x^2}{2}$$

(for $x>0$ proven by using monotonic function) we have

$$-a\sum_{i=1}^n\frac{1}{n}\le\ln a_n\le-a\sum_{i=1}^n\frac{1}{n}+\frac{a^2}{2}\sum_{i=1}^n\frac{1}{n^2}\le-a\sum_{i=1}^n\frac{1}{n}+C$$

(where $C=\frac{a^2}{2}\sum_{i=1}^\infty\frac{1}{n^2}$) Then using the fact that:

$$\ln(n+1)=\int_1^{n+1}\frac{1}{x}dx\leq\sum_{i=1}^n\frac{1}{n}\leq1+\int_1^{n}\frac{1}{x} dx=1+\ln(n)$$

we have

$$e^{-a(1+\ln n)}\le a_n\le e^{C-a(\ln(n+1))}$$

or

$$e^{-a}n^{-a}\le a_n\le e^C(n+1)^{-a}$$

Since $a_n$ is positive, the series converges if and only if $a>1$

• If $a\le 0$ then from $n\ge |a|$ we have $|a_n|$ increases and doesn't change sign so $\lim_{n\rightarrow\infty}\sup |a_n|\ne 0$ then the sum diverges.