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Finding $$\lim_{n\rightarrow \infty}n^{-n^2}\bigg\{(n+1)\bigg(n+\frac{1}{2017}\bigg)\bigg(n+\frac{1}{2017^2}\bigg)\cdots\cdots \cdots \bigg(n+\frac{1}{2017^{n-1}}\bigg)\bigg\}$$

My Try: Assume $$l=\lim_{n\rightarrow \infty}n^{-n^2}\prod^{n}_{r=1}\cdot\bigg(n+\frac{1}{2017^{r-1}}\bigg)$$

$$\ln (l) = -\lim_{n\rightarrow \infty}n^2\ln(n)+\lim_{n\rightarrow \infty}\sum^{n}_{r=1}\ln\bigg(n+\frac{1}{2017^{r-1}}\bigg)$$

could some help me how to solve it, thanks

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    $\begingroup$ Seems that the first expression is wrong? Should the $n^{-n^2}$ be to the left of the $\Pi$? $\endgroup$ – user202729 Dec 12 '17 at 11:28
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Each term in parentheses is not larger than $n+1$. So $l$ is not greater than the limit of $n^{-n^2}(n+1)^{n}$, which is zero.

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  • $\begingroup$ There is a typo. There are $n$ terms in the product so you should have $(n+1)^n$ instead of $(n+1)^{n+1}$ but this typo does not make any difference to the answer. Also "smaller" should be replaced with "not greater" because $l$ is equal to $0$. +1 for a simple approach to a complicated looking problem. $\endgroup$ – Paramanand Singh Dec 12 '17 at 12:57
  • $\begingroup$ Thanks,but I think that there is no typo. Notice that besides the 2017 terms there is an additional $n+1$. Corrected the "smaller". $\endgroup$ – YZS Dec 13 '17 at 15:23
  • $\begingroup$ you can check again. the number of terms in product is $n$ including the first factor $(n +1)$. $\endgroup$ – Paramanand Singh Dec 13 '17 at 16:13
  • $\begingroup$ Oh right I see! I must be getting old. At least my argument is still correct even with this typo -:) $\endgroup$ – YZS Dec 13 '17 at 21:51
  • $\begingroup$ Yes your argument was correct even with the typo (the term $n^{-n^{2}}$ dominates other factor and brings the whole thing down to zero) and that's why you already got an upvote from my end. $\endgroup$ – Paramanand Singh Dec 14 '17 at 2:14
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My reasoning:

Leave the productory, since you have to be careful about sums and limits inside sums and so on.

The major term of the productory will be $n^n$. The others are neglected by the dominating term $n^{-n^2}$.

Hence you are left with $n^{-n^2 + n}$ .

Which goes to $0$

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  • $\begingroup$ Too lazy to write $\lim$. It's understood. $\endgroup$ – Turing Dec 12 '17 at 11:36
  • $\begingroup$ Actually it is not. $\endgroup$ – Did Dec 12 '17 at 12:38
  • $\begingroup$ @Did The productory major term will be $n^n$, the other can be neglected due to the limit. Hence one remains basically with $n^{-n^2 + n}$ which goes as $n^{-n^2}$ which is zero. $\endgroup$ – Turing Dec 12 '17 at 12:40
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    $\begingroup$ No. Please be aware that using "identities" such as $$\lim_{n\to\infty}\sum_{k=1}^nx_{k,n}=\sum_{k=1}^\infty\lim_{n\to\infty}x_{k,n}$$ is a sure way to reach chaos. $\endgroup$ – Did Dec 12 '17 at 12:44
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    $\begingroup$ Yes, "that". If you have a sound proof, simply post it. (Unrelated: FYI, "Productories" does not seem to be an English word.) $\endgroup$ – Did Dec 12 '17 at 12:49

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