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



There is a second part of question when:

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

please help me out.

New contributor
Daksh Verma is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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

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


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

Can you proceed ?


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 $\frac{2999-30 \log (15)}{3000}=0.9726$.


Your Answer

Daksh Verma is a new contributor. Be nice, and check out our Code of Conduct.

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.