What is the sum notation equivalent to this series?

Question

I tried to solve this question

Find radius of convergence of the series
$$1+ \frac{a.b}{1.c}z + \frac{a(a+1)b(b+1)}{1.2...c(c+1)}z^2+.... $$

The book answer is 1

I cant obtain the nth term in the series ?

Answer 1

OK, here is my best guess based on your notation (I think this is correct):

It goes from $a$ to $a(a + 1)$ and then, presumably to $a(a + 1)(a + 2)$, etc. and likewise for $b$. So that's simple to write--it's part of a factorial:

$$ a(a + 1)(a + 2)...(a + n) = \frac{(a + n)!}{(a - 1)!} $$

Then, again, likewise for $b$ and $c$ (except $c$ is on the bottom), then, in addition, you have an $n!$ on the bottom (with the $c$):

$$ \frac{(c - 1)!}{(a - 1)!(b - 1)!}\sum_{n = 0}^\infty \frac{(a - 1 + n)!(b - 1 + n)!}{n!(c - 1 + n)!}z^n $$

(note that when $i = 0$, the factorials give $(a - 1)!$, $(b - 1)!$, and $(c - 1)!$ which all cancel with the coefficient on the outside)

To find the convergence, you would just use the ratio test:

$$ \frac{a_{n + 1}}{a_n} = \frac{(a + n)(b + n)}{(n + 1)(c + n)}z $$

Take the limit as $n \rightarrow \infty$:

$$ \lim_{n \rightarrow \infty} \frac{(a + n)(b + n)}{(n + 1)(c + n)}z = \lim_{n\rightarrow \infty}\frac{n^2}{n^2}z = z $$

This zeries definitely converges when $|z| < 1$ (by the ratio test). However, you now need go back and check $z = 1$ and $z = -1$. I think it would diverge here but it may depend on the value of $a$, $b$, and $c$ (if the coefficients go to zero, then it converges when $z = -1$ by the alternating series test).

Answer 2

Hint: $$a_1=1\text{ and }\\ a_{n\forall n\geq2}=\dfrac{(a+n-1)!(b+n-1)!(c-1)!}{(n-1)!(a-1)!(b-1)!(c-n+1)!}z^{n-1},\text{ assuming $1.2...c(x-1)=1\times2\times c\times (x-1)$.}$$

Answer 3

In terms of rising factorial powers (Knuth's term/notation): $$ x^{\overline{r}} = x (x + 1) \cdots (x + r - 1) $$ your sum seems to be: $$ S(z) = \sum_{n \ge 0} \frac{a^{\overline{n}} b^{\overline{n}}}{n! c^{\overline{n}}} z^n $$ By the ratio test: $$ \frac{\frac{a^{\overline{n + 1}} b^{\overline{n + 1}}}{(n + 1)! c^{\overline{n + 1}}}} {\frac{a^{\overline{n}} b^{\overline{n}}}{n! c^{\overline{n}}}} = \frac{(a + n) (b + n)}{(n + 1) (c + n)} $$ The limit of this is clearly 1, so your serie's convergence radius is 1.