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[tex] \displaystyle \rm\lim_{n \to \infty } \sqrt[n]{ \left( \sum_{k = 1}^{n} \frac{ {k}^{2} }{ {2k}^{2} - 2nk + {n}^{2} } \right) \left( \sum_{k = 1}^{n} \frac{ {k}^3}{ {3k}^{2} - 3nk + {n}^{2} } \right)} [/tex]​ ​

Sagot :

Rewrite the sums as

[tex]\displaystyle S_2 = \sum_{k=1}^n \frac{k^2}{2k^2 - 2nk + n^2} = \sum_{k=1}^n \frac{\frac{k^2}{n^2}}{\frac{2k^2}{n^2} - \frac{2k}n + 1}[/tex]

and

[tex]\displaystyle S_3 = \sum_{k=1}^n \frac{k^2}{3k^2 - 3nk + n^2} = \sum_{k=1}^n \frac{\frac{k^2}{n^2}}{\frac{3k^2}{n^2} - \frac{3k}n + 1}[/tex]

Now notice that

[tex]\displaystyle \lim_{n\to\infty} \frac{S_2}n = \int_0^1 \frac{x^2}{2x^2 - 2x + 1} = \frac12[/tex]

and

[tex]\displaystyle \lim_{n\to\infty} \frac{S_3}n = \int_0^1 \frac{x^2}{3x^2 - 3x + 1} = \frac{9 + 2\pi\sqrt3}{27}[/tex]

and the important point here is that [tex]\frac{S_2}n[/tex] and [tex]\frac{S_3}n[/tex] converge to constants. For any real constant a, we have

[tex]\displaystyle \lim_{n\to\infty} \frac{\ln(an)}n = 0[/tex]

Rewrite the limit as

[tex]\displaystyle \lim_{n\to\infty} \sqrt[n]{S_2 \times S_3} = \lim_{n\to\infty} \exp\left(\ln\left(\sqrt[n]{S_2 \times S_3}\right)\right) \\\\ = \exp\left(\lim_{n\to\infty} \frac{\ln(S_2) + \ln(S_3)}n\right) \\\\ = \exp\left(\lim_{n\to\infty} \frac{\ln\left(n \times \frac{S_2}n\right) + \ln\left(n \times \frac{S_3}n\right)}n\right)[/tex]

Then

[tex]\displaystyle \lim_{n\to\infty} \sqrt[n]{S_2 \times S_3} = e^0 = \boxed{1}[/tex]

A plot of the limand for n = first 1000 positive integers suggests the limit is correct, but convergence is slow.

View image LammettHash
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