prove that $\prod_{j=1}^{n} (1+\frac{a}{j})=Cn^a(1+O(1/n))$ as $n \to \infty$

Question

I want to prove that $\displaystyle \prod_{j=1}^{n} (1+\frac{a}{j})=Cn^a(1+O(1/n))$ as $n \to \infty$. where $C$ is a constant independent of $n$. For $a=1$, it is simply seen to be true. How can I prove this asymptotic relation for arbitrary $a$? I would appreciate any help/hint/answers.

Thanks,

Answer 1

Here is an elementary proof that $\prod_{j=1}^n (1+\frac{a}{j}) =\Theta(n^a) $.

You have to work harder to get $Cn^a(1+O(1/n)) $. Maybe show that $Cn^a(1+O(1/n))(1+\frac{a}{n+1}) =C(n+1)^a(1+O(1/n)) $.

Thinking about this, $\dfrac{(n+1)^a}{n^a} =(1+\frac1{n})^a =1+\frac{a}{n}+O(1/n^2) =1+\frac{a}{n+1}+O(1/n^2) $ since $\frac{a}{n}-\frac{a}{n+1} =\frac{a}{n(n+1)} =O(1/n^2) $. Not sure how rigorous this is.

So I'll proceed with my original answer.

As clark suggested, $f(n, a) =\prod_{j=1}^n (1+\frac{a}{j}) =e^{\sum_{j=1}^n \ln ( (1+\frac{a}{j}) )} =e^{g(n, a)} $.

We have $x-x^2/2 < \ln(1+x) < x $ for $x > 0$ (proof if needed) so

$\begin{array}\\ g(n, a) &\lt \sum_{j=1}^n \frac{a}{j}\\ &=a \sum_{j=1}^n \frac{1}{j}\\ &=a H_n\\ &<a (\ln(n)+1)\\ \text{so}\\ f(n, a) &<e^{a (\ln(n)+1)}\\ &=n^ae^{a}\\ \end{array} $

Similarly,

$\begin{array}\\ g(n, a) &\gt \sum_{j=1}^n (\frac{a}{j}-\frac{a^2}{2j^2})\\ &=a \sum_{j=1}^n \frac{1}{j}-\frac{a^2}{2}\sum_{j=1}^n \frac{1}{j^2})\\ &>a H_n-a^2\\ &>a \ln(n)-a^2\\ \text{so}\\ f(n, a) &>e^{a \ln(n)-a^2)}\\ &=n^ae^{-a^2}\\ \end{array} $

Answer 2

A quick implementation of clark's suggestion:

From the Taylor series of $\log$ we get

$$ \log\!\left(1 + \frac{a}{j}\right) - \frac{a}{j} \sim -\frac{a^2}{2j^2} $$

as $j \to \infty$, and so the series

$$ C := \sum_{j=1}^{\infty} \left[\log\!\left(1 + \frac{a}{j}\right) - \frac{a}{j}\right] $$

converges. In fact,

$$ \sum_{j=1}^{n} \left[\log\!\left(1 + \frac{a}{j}\right) - \frac{a}{j}\right] = C + O\!\left(\frac{1}{n}\right). $$

Thus

$$ \begin{align} \log \prod_{j=1}^{n} \left(1 + \frac{a}{j}\right) &= \sum_{j=1}^{n} \log\!\left(1 + \frac{a}{j}\right) \\ &= a\sum_{j=1}^{n} \frac{1}{j} + C + O\!\left(\frac{1}{n}\right) \\ &= a\log n + a\gamma + C + O\!\left(\frac{1}{n}\right), \end{align} $$

where in the last line we used the well-known approximation $H_n = \log n + \gamma + O(1/n)$.

Finally we get

$$ \prod_{j=1}^{n} \left(1 + \frac{a}{j}\right) = n^a e^{a\gamma + C + O(1/n)} = n^a e^{a\gamma + C} \left[ 1 + O\!\left(\frac{1}{n}\right)\right] $$

as $n \to \infty$.