I was trying to find a proof for Raabe's Ratio test: If $x_n$ is a positive sequence of real numbers, if $\lim\limits_{n \to \infty } n \left(\frac{x_n}{x_{n+1}}-1 \right) >1$ then $\sum\limits_{n=1}^ \infty x_n$ converges.
So I tried to make this argument: Since $\lim\limits_{n \to \infty } n \left(\frac{x_n}{x_{n+1}}-1 \right) >1 $, then there exists $N \in \mathbb{N} $ such that for all $n\ge N$, $n \left(\frac{x_n}{x_{n+1}}-1 \right)>a >1$ for some real $a$.
$$\frac{x_n}{x_{n+1}}>\frac{a+n}{n}$$
$$x_{n+1} < x_n\left(1-\frac{a}{a+n}\right)$$
i.e. $x_{N+1}< x_N\left(1-\frac{a}{a+N}\right), $ $x_{N+2}<x_N\left(1-\frac{a}{a+N}\right)\left(1-\frac{a}{a+N+1}\right) $ and so on.
i.e. $$\sum\limits_{n=N }^ \infty x_n <x_N+x_N \sum_{n=1}^\infty \prod_{k=0}^{n-1} \left(1- \frac{a}{a+N +k} \right)$$
so I tried to prove that for all $a>1$, the sum $\displaystyle \sum_{n=1}^\infty \prod_{k=0}^{n-1} \left(1- \frac{a}{a+N +k} \right)$ converges, which would be a proof of Raabe's Ratio test. The only problem is that I couldn't find a proof and I don't know if this sum always converge or not.
I want to ask for an elementary proof using tests of convergence or epsilon proofs without using special functions like gamma.
Best Answer
Multiply your series by $$\Gamma(N)\Gamma(a)/ \Gamma(a+N)$$ and your series becomes $$\sum_{n=1}^{\infty}\frac{\Gamma(a)\Gamma(N)\prod_{k=0}^{n-1} (N+k)}{\Gamma(a+N)\prod_{k=0}^{n-1} (a+N+k)}=\sum_{n=1}^{\infty} B(N+n, a) =\int_0^1 (1-x)^{a-1}\sum_{n=1}^\infty x^{N+n-1} \mathrm d x = \int_{0}^1(1-x)^{a-2} x^{N} \mathrm dx = B(N+1, a-1)$$
So your series is equal to
$$\frac{\Gamma(a-1)\Gamma(N+1)}{\Gamma(a)\Gamma(N)} = \frac{N}{a-1}$$
Another way of doing it is the following: let $u_n = \prod_{k=1}^{n}\frac{k}{a+k}$ and $v_n = \prod_{k=1}^n \frac{k}{a-1+k}$ you can prove that,
$$\lim\limits_{n\to \infty} v_n =0$$
Indeed using $\ln (1+x) \ge x - x^2/2$,
\begin{align} \ln v_n &= -\sum_{k=1}^{n} \ln \left(1 + \frac{a-1}{k}\right)\\ &\le - \sum_{k=1}^{n} \frac{a-1}{k} + \frac12\sum_{k=1}^n \frac{(a-1)^2}{k^2} \to -\infty \end{align}
Since $\sum\frac1k$ diverges and $\sum\frac1{k^2}$ converges.
Now you can prove that, \begin{align} v_n - v_{n+1} &= \prod_{k=1}^n \frac{k}{a-1+k} - \prod_{k=1}^{n+1} \frac{k}{a-1+k}\\ &= \left(\prod_{k=1}^n \frac{k}{a-1+k}\right)\left(1-\frac{n+1}{a-1+n+1}\right)\\ &= \left(\prod_{k=1}^n \frac{k}{a-1+k}\right)\left(\frac{a-1}{a+n}\right) = u_n \end{align}
On the other hand,
\begin{align} \sum_{n=1}^{\infty} \prod_{k=0}^{n-1}\left(1-\frac{a}{a+N+k}\right) &= \sum_{n=1}^{\infty} \prod_{k=0}^{n-1}\frac{N+k}{a+N+k}\\ &= \sum_{n=1}^{\infty} \prod_{k=N}^{n+N-1}\frac{k}{a+k}\\ &= \sum_{n=N}^{\infty} \prod_{k=N}^{n} \frac{k}{a+k}\\ &= \frac1{\prod_{k=1}^{N-1} \frac{k}{a+k}}\sum_{n=N}^{\infty} u_n\\ &= \left(\prod_{k=1}^{N-1} \frac{a+k}{k}\right)\sum_{n=N}^{\infty} (v_n - v_{n+1})\\ &= \left(\prod_{k=1}^{N-1} \frac{a+k}{k}\right) v_N \end{align}
Can you now finish the proof?