# summation of a infinite series

i want to know if this infinite sum $\sum_{n=1}^{\infty}$ $\frac{a(a+1)(a+2)......(a+n-1)}{b(b+1)(b+2)......(b+n-1)}$ converges or diverges ? where a>0 and b>a+1.

if the sum converges what is the sum i.e where it will converge? i need a concrete explanation

i found some inequalities $\frac{a(a+1)}{b(b+1)}$ <$\frac{a}{b+1}$ and $\frac{a(a+1)(a+2)}{b(b+1)(b+2)}$< $\frac{a}{b+2}$ .......and continuing $\frac{a(a+1)....(a+n-1)}{b(b+1)....(b+n-1)}$ < $\frac{a}{b+n-1}$

this inequalities can be useful for this problem

• @tired Case $a=1, b=3$ is easy to solve, in this case the sum exists and equals $1$. So the answer "the sum always diverges" is obviously wrong. Nov 28, 2016 at 11:29

Since Hamou has answered the question of convergence, I'll answer the question about the sum of the series. For $b - a > 1$, the sum of the series is $$-1 + \frac{\Gamma(b)\Gamma(b-a-1)}{\Gamma(b-a)\Gamma(b-1)}$$

Indeed, Abel's continuity theorem gives

$$\sum_{n = 1}^\infty \frac{a(a+1)\cdots (a + n-1)}{b(b + 1)\cdots (b + n - 1)} = \lim_{x\to 1^{-}} \sum_{n = 1}^\infty \frac{a(a+1)\cdots (a+n-1)}{b(b+1)\cdots (b+n-1)}x^n$$

For $\lvert x \rvert < 1$,

$$\sum_{n = 0}^\infty \frac{a(a+1)\cdots(a+n-1)}{b(b+1)\cdots(b+n-1)}x^n = \frac{\Gamma(b)}{\Gamma(a)}\sum_{n = 0}^\infty \frac{\Gamma(a+n)}{\Gamma(b+n)}x^n = \frac{\Gamma(b)}{\Gamma(b-a)\Gamma(a)}\sum_{n = 1}^\infty \frac{\Gamma(b-a)\Gamma(a+n)}{\Gamma(b+n)}x^n$$

The quotients $\Gamma(b-a)\Gamma(a+n)/\Gamma(b+n)$ are represented by Beta integrals

$$\int_0^1 (1 - t)^{b-a-1}t^{a+n-1}\, dt$$

Thus

$$\sum_{n = 0}^\infty \frac{\Gamma(b-a)\Gamma(a+n)}{\Gamma(b+n)}x^n = \int_0^1 \sum_{n = 0}^\infty (tx)^n(1-t)^{b-a-1}t^{a-1}\, dt = \int_0^1 (1 - tx)^{-1}(1-t)^{b-a-1}t^{a-1}\, dt$$

This gives the integral representation

$$\sum_{n = 0}^\infty \frac{a(a+1)\cdots(a+n-1)}{b(b+1)\cdots(b+n-1)}x^n = \frac{\Gamma(b)}{\Gamma(b-a)\Gamma(a)}\int_0^1 (1 - tx)^{-1}(1-t)^{b-a-1}t^{a-1}\, dt$$

Taking the limit as $x\to 1^{-}$ results in

$$\frac{\Gamma(b)}{\Gamma(b-a)\Gamma(a)} \int_0^1 (1 - t)^{b-a-2}t^{a-1}\, dt = \frac{\Gamma(b)}{\Gamma(b-a)\Gamma(a)}\frac{\Gamma(b-a-1)\Gamma(a)}{\Gamma(b-1)} = \frac{\Gamma(b)\Gamma(b-a-1)}{\Gamma(b-a)\Gamma(b-1)}$$

So

$$\sum_{n = 1}^\infty \frac{a(a+1)\cdots(a+n-1)}{b(b+1)\cdots(b+n-1)} = -1 + \sum_{n = 0}^\infty \frac{a(a+1)\cdots(a+n-1)}{b(b+1)\cdots(b+n-1)} = -1 + \frac{\Gamma(b)\Gamma(b-a-1)}{\Gamma(b-a)\Gamma(b-1)}$$

Let $u_n=\dfrac{a(a+1)\ldots(a+n-1)}{b(b+1)\ldots(b+n-1)}$

We have $\dfrac{u_{n+1}}{u_n}=1-\dfrac{b-a}{n}+o(\dfrac{1}{n})$

Hence by Raabe Duhamel :

If $b-a>1$ the infinite sum converge

If $b-a<1$ the infinite sum diverge.

• The question was "what is the sum" as well. Also, you wrote wrong expression for $u_n$ Nov 28, 2016 at 12:07
• what is raabe Duhamel. ? Nov 28, 2016 at 17:54
• See the link. . Nov 29, 2016 at 9:27