determine for which value $\alpha $ the sequence
$a_n = (1- (\frac{1}{n})^\alpha ) ( 1-(\frac{2}{n})^\alpha)......(1-(\frac{n-1}{n})^\alpha )$ and $n\ge 2$ converges
i think when $\alpha = 0 $ definitely it will converges to $ o$
afterthat when $\alpha \neq 0$ then what happen.??? as i can not able to proceed further.....................................................................
pliz help me
For $\alpha>0$, \begin{align*} \log a_{n}=\dfrac{(1/n)\displaystyle\sum_{k=0}^{n-1}\log\left(1-(k/n)^{\alpha}\right)}{1/n}, \end{align*} we see that \begin{align*} \dfrac{1}{n}\sum_{k=0}^{n-1}\log(1-(k/n)^{\alpha})\rightarrow\int_{0}^{1}\log(1-x^{\alpha})dx<0, \end{align*} so $\log a_{n}\rightarrow-\infty$, so $a_{n}\rightarrow 0$.
Now consider $\alpha<0$, write $\alpha=-p$, $p>0$, we first deal with $(-1)^{n}a_{n}$, then \begin{align*} \log(-1)^{n}a_{n}=\dfrac{(1/n)\displaystyle\sum_{k=0}^{n-1}\log\left(1/(k/n)^{p}-1\right)}{1/n}, \end{align*} we see that \begin{align*} \dfrac{1}{n}\sum_{k=0}^{n-1}\log(1/(k/n)^{p}-1)\rightarrow\int_{0}^{1}\log\left(\dfrac{1}{x^{p}}-1\right)>0, \end{align*} so $\log(-1)^{n}a_{n}\rightarrow\infty$ and hence $(-1)^{n}a_{n}\rightarrow\infty$. If it were $a_{n}\rightarrow L$ for some real number $L$, then $(-1)^{2k}a_{2k}\rightarrow L$, a contradiction.