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I would like study the convergence of the following sequence:

$ u_{n}=\sqrt [n]{\frac{(a+1)(a+2)...(a+n)}{n!}} $

where $a>0$

We have:

$ \ln(u_{n})=\frac{1}{n}\sum_{k=1}^n \ln(1+\frac{a}{k})$

And : $ \ln(u_{n+1})-\ln(u_{n})=\frac{1}{n+1}\sum_{k=1}^{n+1} \ln(1+\frac{a}{k})-\frac{1}{n}\sum_{k=1}^n \ln(1+\frac{a}{k})$

So I have to study the convergence of $ \sum \ln(u_{n+1})-\ln(u_{n})$

Using integrals: $ \sum_{k=1}^n \ln(1+\frac{a}{k}) \sim a\ln(n)$

Thus: $ \ln(u_{n+1})-\ln(u_{n})= \frac{1}{n}a\ln(n)-\frac{1}{n}a\ln(n)+o(\frac{\ln(n)}{n})=o(\frac{\ln(n)}{n}) $

which is not enough to determine the convergence or divergence of $ \sum \ln(u_{n+1})-\ln(u_{n})$

So how can I find a more precise approximation of $ \ln(u_{n+1})-\ln(u_{n})$?

Or is there a simple equivalent of $ \prod_{k=1}^n {(a+k)}$ ?

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    Have you tried using the fact that $\ln n!\approx (n-1)\ln n$ with great accuracy? This is known as Stirling's approximation.2012-02-26

3 Answers 3

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Another proof, which uses the $\text{GM} \le \text{AM}$ inequality.

We have

$1 \le \sqrt[n]{\frac{(a+1)}{1}\cdot \frac{(a+2)}{2} \cdots \frac{(a+n)}{n}} \le \frac{1}{n}\sum_{k=1}^{n} (1 + \frac{a}{k}) = 1 + \frac{1}{n} \sum_{k=1}^{n} \frac{a}{k}$

Since $\frac{a}{n} \to 0$, we have that $\frac{1}{n} \sum_{k=1}^{n} \frac{a}{k} \to 0$ and so your sequence converges to $1$.

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    @PeterT.off: Thank you!2012-02-26
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Just a note on evaluating the limit of your sequence:

You can compute the limit by using the fact that if $\lim\limits_{n\rightarrow\infty}{a_{n+1}\over a_n} $ exists, then so does $\lim\limits_{n\rightarrow\infty}{\root n\of {a_n}}$ and the two limits are equal.

Set $a_n={(a+1)(a+2)\cdots(a+n)\over n!}$. Then a simple computation shows that $\lim\limits_{n\rightarrow\infty}{a_{n+1}\over a_n}=1 $. So, $\lim\limits_{n\rightarrow\infty}{\root n\of{ a_n}} =\lim\limits_{n\rightarrow\infty}{\root n\of {(a+1)(a+2)\cdots(a+n)\over n!}} =1$.

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    @PlaneChon-Ju By the way, Ragib's answer is much simpler; and you more or less had the same solution waiting already. Just use the approximation in your fifth displayed equation.2012-02-26
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Since $ \log(1+x) \leq x$ and $\displaystyle \sum_{k=1}^n \frac{1}{k} \sim \log n, $ $ \ln(u_{n})=\frac{1}{n}\sum_{k=1}^n \log \left(1+\frac{a}{k}\right) \leq \frac{a}{n}\sum_{k=1}^n\frac{1}{k} \to 0$

thus $u_n \to 1.$