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