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In an exam I have been asked to discuss the convergence of a series regarding a parameter $d$. Here's the following : $\sum_{n=0}^\infty \frac{(n!)^d}{(d\cdot n)!}$

The answer is that this series converges for $d \geq 2$. I totally understand that if $d \leq 1$, the series will not converge but I am blocked while trying to use the d'Alembert or Cauchy's rules.

Can somebody give me a tip ?

  • 1
    Is that $(d\cdot n)!$ or $d\cdot (n!)$ in the denominator? – 2017-01-04
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    unfortunately it is (d*n)! I thought i wrote it like that sorry – 2017-01-04
  • 0
    Is $d$ an integer or?If not how do you define $(d\cdot n)!$ for non integers? – 2017-01-04
  • 0
    Yes d is an integer – 2017-01-04

2 Answers 2

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Let $f(n)$ be the general term of the series. Then, we have:

$$\frac{f(n+1)}{f(n)} = \frac{(n+1)^d}{(dn+d)(\cdots)(dn +1)}\le \left( \frac{n+1}{dn + 1}\right)^d \to \frac1{d^d}$$

Thus:

$$\limsup \frac{f(n+1)}{f(n)} \le \frac1{d^d} < 1 \text{ for $d\ge 2$}$$

So the series converges (using the ratio test).

  • 0
    I just ask myself in this example why the factorial of the numerator disappeared (I might have misunderstood) – 2017-01-04
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    @UlysseTouchais things cancel out. Try to work it out. – 2017-01-04
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Stirling approximation yields $\displaystyle \frac{(n!)^d}{(d\cdot n)!}\sim K\frac{n^{\frac{d-1}2}}{(d^d)^n}$ where $K=\frac{(\sqrt{2\pi})^{d-1}}{\sqrt d}$

When $d>1$, $d^d>1$ and $\displaystyle \frac{n^{\frac{d-1}2}}{(d^d)^n} = O\left( \frac{1}{(d^d)^{n/2}}\right)$ and the series converges.

When $d=1$, $\displaystyle \frac{n^{\frac{d-1}2}}{(d^d)^n} = 1$ and the series diverges.

When $d<1$, $\displaystyle \frac{n^{\frac{d-1}2}}{(d^d)^n}\to \infty$ and the series diverges trivially.