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The n-th term is given by $$a_{n}:=\frac{n}{\sqrt[3]{n^{6}-1}}+\frac{n}{\sqrt[3]{n^{6}-2}}+...+\frac{n}{\sqrt[3]{n^{6}-n-2}}$$ then using the squeeze theorem I find that: $$\frac{(n+1)n}{\sqrt[3]{n^{6}-n-2}}\leq \frac{n}{\sqrt[3]{n^{6}-1}}+\frac{n}{\sqrt[3]{n^{6}-2}}+...+\frac{n}{\sqrt[3]{n^{6}-n-2}}\leq\frac{(n+1)n}{\sqrt[3]{n^{6}-1}}$$ for every $n\in \mathbb{N}\setminus\left\{1\right\}$, and since the limits of both the left hand side and right hand side sequence are equal to one, I find that $$\lim _{n\rightarrow \infty }\frac{n}{\sqrt[3]{n^{6}-1}}+\frac{n}{\sqrt[3]{n^{6}-2}}+...+\frac{n}{\sqrt[3]{n^{6}-n-2}}=1$$ The next part of the problem is to find $\lim _{n\rightarrow \infty }(a_{n})^{2}$, which is the same as $(\lim _{n\rightarrow \infty }a_{n})^{2}$, and thus has the same solution, namely 1. Is this correct? The next task is to find $\lim _{n\rightarrow \infty }(a_{n})^{n}$, which, by the same logic as stated above, equals one, too. The problem is that this answer isn't given in the answer sheet. One of the possible solutions to this is $e^{2}$, which according to my wolfram calculus app is correct. But how to get there?

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    $\begingroup$ To illustrate the issue in a form you may be familiar with, if $a_n=1+\frac1n$ then the limit of $a_n$ is $1$ but the limit of $a_n^{n}$ is $e^1$. I suspect you were asked for $\lim _{n\rightarrow \infty }(a_{n}^{n})$ rather than $(\lim _{n\rightarrow \infty }a_{n})^{n}$, which makes no sense $\endgroup$ – Henry Dec 13 '14 at 20:38
  • $\begingroup$ Yeah, now I see. I will edit the question, but one more thing, isn't that just the same thing when the sequence converges to one? $\endgroup$ – Shemafied Dec 13 '14 at 21:46
  • $\begingroup$ No it is not the same, as my example shows $\endgroup$ – Henry Dec 13 '14 at 23:16
  • $\begingroup$ Alright, now I see why. But, is my first assumption correct, for the power of two? And what steps do you recommend for solving it with the power of n? $\endgroup$ – Shemafied Dec 13 '14 at 23:19
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There are $n+2$ terms in the definition of $a_n$, and the terms in the sum are increasing, so you should have

$$\frac{(n+2)n}{\sqrt[3]{n^{6}-1}}\leq a_n \leq \frac{(n+2)n}{\sqrt[3]{n^{6}-n-2}}$$

but your squeeze argument would still work to say $a_n \to 1$. Your argument would also work for $a_n^2 \to 1$, but it would not work $a_n^n$.

Personally I would suggest trying for a series expansion of $\frac{(n+2)n}{\sqrt[3]{n^{6}-n-2}}$ and $\frac{(n+2)n}{\sqrt[3]{n^{6}-1}}$. If you find it difficult directly, then you could look at the series expansion of their cubes by polynomial division, which both start $1+6n^{-1}+12n^{-2}+8n^{-3}+\cdots$, so $\lim a_n^{3n} = \lim \left(1+\frac6n\right)^n=e^6$ and taking cube roots will give you $\lim a_n^{n}$.

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  • $\begingroup$ I would say that there are n+1 terms. How did you get that number anyway? $\endgroup$ – Shemafied Dec 16 '14 at 14:46
  • $\begingroup$ Suppose $n=2$. Then $n^6-1=63$ and $n^6-n-2=60$ and you have $\dfrac{2}{\sqrt{63}}+\dfrac{2}{\sqrt{62}}+\dfrac{2}{\sqrt{61}}+\dfrac{2}{\sqrt{60}}$, which is the sum of $4$ (i.e. $n+2$) terms $\endgroup$ – Henry Dec 16 '14 at 14:49
  • $\begingroup$ Oh, sorry that I reply so late, but I had already figured it out on my own, and even found an easier way to calculate this limit. Do you wanna see? $\endgroup$ – Shemafied Dec 16 '14 at 15:01
  • $\begingroup$ You can if you wish edit your question or post your own answer. $\endgroup$ – Henry Dec 16 '14 at 15:02
  • $\begingroup$ There, I added an answer. $\endgroup$ – Shemafied Dec 16 '14 at 15:15
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$$\lim _{n\rightarrow \infty }(a_{n})^{n}=\lim \left ( \frac{n^{2}}{\sqrt[3]{n^{6}-1}} +\frac{2n}{\sqrt[3]{n^{6}-1}}\right )^{n}=\lim \left ( 1+ \frac{1}{\frac{\sqrt[3]{n^{6}-1}}{2n}}\right )^{\frac{\sqrt[3]{n^{6}-1}}{2n}\cdot \frac{2n^{2}}{\sqrt[3]{n^{6}-1}}}=e^{2}$$

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