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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

1 Answers1

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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.

user284331
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  • im not getting the last line,,,If it were $a_{n}\rightarrow L$ for some real number $L$, then $(-1)^{2k}a_{2k}\rightarrow L$, a contradiction....can u elaborate more –  Apr 05 '18 at 06:11
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    Because $((-1)^{2k}a_{2k})=(a_{2k})$ which is a subsequence of $(a_{n})$, must have the same limit point, so $(-1)^{2k}a_{2k}\rightarrow L$, but we have on the other hand that $(-1)^{2k}a_{2k}\rightarrow\infty$ as a subsequence of $((-1)^{n}a_{n})$. – user284331 Apr 05 '18 at 06:13
  • that mean at$ \alpha = -1 $ it will diverges,,, tell me now the value of $ \alpha $,,at which sequence will converges.. –  Apr 05 '18 at 06:17
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    Then only $\alpha=0$ goes through. – user284331 Apr 05 '18 at 06:18
  • that mean the given sequence ${a_n}$ converges only at $ \alpha = 0$..Am i right ?? –  Apr 05 '18 at 06:21
  • thanks a lots @user284331 –  Apr 05 '18 at 06:23
  • Sorry, for that $\alpha>0$, the sequence also converges, the limit is zero. So for all $\alpha\geq 0$, the sequence converges. – user284331 Apr 05 '18 at 06:43
  • @user284331 In your proof you use the fact 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} But this is not so obvious. What is indeed TRUE is that \begin{align} \dfrac{1}{n}\sum_{k=0}^{n-1}f(k/n)\rightarrow\int_{0}^{1}f(x),dx, \end{align} provided that $f$ is CONTINUOUS in $[0,1]$ . Unfortunately, $\log(1-x^a)$ is NOT continuous. Not even bounded! – Yiorgos S. Smyrlis Apr 05 '18 at 07:26
  • Yes, that is wrong? – user284331 Apr 05 '18 at 07:28
  • @user284331 You should show it! – Yiorgos S. Smyrlis Apr 05 '18 at 07:30
  • Ah... I missed that, but intuitively, because $\alpha>0$ and $x\in[0,1)$, the function $\log(1-x^{\alpha})$ has only negative sign, and it is improper integrable, things should be not that bad, but let me think how to properly fill in the details. – user284331 Apr 05 '18 at 07:34
  • Fortunately, this post addresses this issue. – user284331 Apr 05 '18 at 08:29