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

  • 1
    $\begingroup$ 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

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.

  • $\begingroup$ 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

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 $0$

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.