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If: $$a_n=(1+1/n^2) + (1+2/n^2)^2 + (1+3/n^2)^3+\cdots+(1+n/n^2)^n =\sum_{k=1}^n \left(1 + \frac{k}{n^2} \right)^k $$

find:

$$\lim_{n\rightarrow\infty}(a_n)^{-1/n^2}$$

There is a second part of question when:

$$(a)^{-1/n^2} $$

please help me out.

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  • $\begingroup$ What have you tried? Also, please refer to this link for formatting future questions: math.meta.stackexchange.com/questions/5020/… $\endgroup$ – scoopfaze 2 days ago
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    $\begingroup$ Hey, I suggest to start by noting that the terms in the sum are increasing. Also, what would be the second part of the question? $\endgroup$ – Brightsun 2 days ago
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Let $a_n := \sum_{k=1}^n \left(1 + \frac{k}{n^2} \right)^k.$

Since $\left(1 + \frac{k}{n^2} \right)^k \le 2^k$ for $k=1,...,n$, we get

$$1 \le a_n \le \sum_{k=1}^n 2^k = 2^{n+1}-2 \le 2^{n+1},$$

thus

$$ 1 \le a_n^{1/n^2} \le 2^{\frac{n+1}{n^2}}.$$

Can you proceed ?

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Being myself very lazy, considering $$a_n = \sum_{k=1}^n \left(1 + \frac{k}{n^2} \right)^k$$ just compute the very first values and notice that it is almost a straight line.

So, assuming $a_n=\alpha+\beta\, n$ $$A_n=(\alpha +\beta n)^{-\frac{1}{n^2}}\implies \log(A_n)={-\frac{1}{n^2}}\log(\alpha +\beta n)$$ Expanding as Taylor series $$\log(A_n)={-\frac{1}{n^2}}\left(\log (\beta)+\log \left({n}\right)+\frac{\alpha}{\beta n}+O\left(\frac{1}{n^2}\right)\right)$$ $$A_n=e^{\log(A_n)}=1-\frac{\log \left({n}\right)+\log (\beta)}{n^2}-\frac{\alpha}{\beta n^3}+O\left(\frac{1}{n^4}\right)$$ Still more lazy, using only $a_1=2$ and $a_2=\frac 72$ gives $\alpha=\frac 12$ and $\beta=\frac 32$.

Using the simplistic material given above, for $n=10$, the exact value is $0.9731$ while the approximation gives $\frac{2999-30 \log (15)}{3000}=0.9726$.

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