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$$\sum_{n=1}^\infty~\frac{n!}{(a+1)(a+2)...(a+n)}$$ Maybe $\lim_{n\to \infty} a_n \ne0$ ?

yes, sorry, $a>0$

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I think you can write the denominator as $\displaystyle\frac{(a+n)!}{a!}$ –  V-Moy May 18 at 12:20
    
What is the condition on $a$? Try to apply one of Raabe or Kummer or Gauß-criterions. –  LutzL May 18 at 12:20
    
Give the ratio test a try. I assume $a$ is a real number, and not a negative integer. –  Gerry Myerson May 18 at 12:20
    
@GerryMyerson You give it a try. When have you ever seen a textbook series exercise with factorials that falls for the ratio test? –  user149090 May 18 at 12:27
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3 Answers 3

up vote 3 down vote accepted

Since $a_n=\binom{a+n}{n}^{-1}$, try to represent it using the difference of two sequence elements when replacing $a$ by $a-1$ or $a+1$, \begin{align} \binom{a-1+n}{n}^{-1}-\binom{a+n}{n+1}^{-1} &= \frac{n!}{(a+n-1)...(a)(a-1)}-\frac{(n+1)!}{(a+n)...(a)(a-1)}\\[0.8em] &=\frac{n!(a+n-(n+1))}{(a+n)...(a)(a-1)}=a_n \end{align} which allows you to examine this as telescoping series.

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The binomial coefficients are also defined for real or complex "numerators". If both entries are non-integer, one can still use beta- and gamma functions. –  LutzL May 18 at 12:32
    
i love this forum so much! Thank a lot –  slmkarta May 18 at 12:34
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I just want to add another method. Not because it is going to provide a cleaner solution, but because the method itself could be useful for other problems.

For dealing with the growth of factorials it is often convenient to use Stirling's

$n!\sim n^ne^{-n}\sqrt{2\pi n}$ as $n\to\infty$.

Then the series is equivalent to the series of

$$\frac{n^ne^{-n}\sqrt{n}a^ae^{-a}\sqrt{a}}{(a+n)^{a+n}e^{-(a+n)}\sqrt{a+n}}=\frac{n^na^a}{(a+n)^{a+n}}\frac{\sqrt{n}\sqrt{a}}{\sqrt{a+n}}$$

We can forget about $\frac{a^a\sqrt{n}\sqrt{a}}{\sqrt{a+n}}$, since it tends to a number different from zero.

We get

$$\frac{1}{(a+n)^a}\frac{1}{(1+a/n)^n}.$$

Since the second fraction also converges to a number different from $0$ we can also forget about it.

Then the series is equivalent to $\frac{1}{n^a}$.

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The difference of reciprocals of binomial polynomials of degree $a-1$ is proportional to the reciprocal of a binomial polynomial of degree $a$; to be precise, $$ \frac1{\binom{n+a}{a}}=\frac{a}{a-1}\left[\frac1{\binom{n+a-1}{a-1}}-\frac1{\binom{n+a}{a-1}}\right]\tag{1} $$ If we sum in $n$, the right side of $(1)$ telescopes. Thus, for $a\gt1$, $$ \begin{align} \sum_{n=1}^\infty\frac1{\binom{n+a}{a}} &=\frac{a}{a-1}\frac1{\binom{a}{a-1}}\\ &=\frac1{a-1}\tag{2} \end{align} $$ For $a=1$, the series diverges since $$ \sum_{n=1}^\infty\frac1{\binom{n+1}{1}}=\sum_{n=1}^\infty\frac1{n+1}\tag{3} $$ diverges.

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