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Sagot :
A possible starting point:
Split up the limit as
[tex]\displaystyle \lim_{n\to\infty} \frac{1 + \sqrt[3]{2} + \cdots + \sqrt[3]{n}}{n^{4/3}} \times \lim_{n\to\infty} \frac1{n \left(\frac1{(an+1)^2} + \frac1{(an+2)^2} + \cdots + \frac1{(an+n)^2}\right)} = 54[/tex]
Consider the first limit,
[tex]\displaystyle \lim_{n\to\infty} \frac{1 + \sqrt[3]{2} + \cdots + \sqrt[3]{n}}{n^{4/3}}[/tex]
Refer to the Stolz-Cesàro theorem, which says
[tex]\displaystyle \lim_{n\to\infty} \frac{a_n}{b_n} = \lim_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}[/tex]
where [tex]a_n[/tex] and [tex]b_n[/tex] are two real sequences, with [tex]b_n[/tex] monotone and divergent. In this case,
[tex]a_n = 1+\sqrt[3]{2}+\sqrt[3]{3}+\cdots+\sqrt[3]{n}[/tex]
[tex]b_n = n^{4/3}[/tex]
Applying S-C, we get
[tex]\displaystyle \lim_{n\to\infty} \frac{\sqrt[3]{n+1}}{(n+1)^{4/3} - n^{4/3}} = \lim_{n\to\infty} \frac{(n+1)^{1/3}}{(n+1)^{4/3} - n^{4/3}}[/tex]
Recalling the difference of cubes identity,
[tex]a^3 - b^3 = (a - b) (a^2 + ab + b^2)[/tex]
we can rewrite the limit as
[tex]\displaystyle \lim_{n\to\infty} \frac{(n+1)^3 + (n+1)^{5/3} n^{4/3} + (n+1)^{1/3} n^{8/3}}{(n+1)^4 - n^4}[/tex]
and dividing uniformly through the limand by (n + 1)³ yields
[tex]\displaystyle \lim_{n\to\infty} \frac{1 + \left(\frac n{n+1}\right)^{4/3} + \left(\frac n{n+1}\right)^{8/3}}{(n+1) - \frac{n^4}{(n+1)^3}}[/tex]
Now,
[tex]n^4 = (n+1)^4 - 4n^3 - 6n^2 - 4n - 1[/tex]
[tex]\implies \dfrac{n^4}{(n+1)^3} = (n+1) - \dfrac{4n^3+6n^2+4n+1}{(n+1)^3}[/tex]
so the denominator in the limit reduces to a degree-1 polynomial with leading coefficient +4. The numerator converges to 1 + 1 + 1 = 3, so this first limit evaluates to
[tex]\displaystyle \lim_{n\to\infty} \frac{1 + \sqrt[3]{2} + \cdots + \sqrt[3]{n}}{n^{4/3}} = \frac34[/tex]
It remains to determine the value of a such that
[tex]\displaystyle \lim_{n\to\infty} \frac1{n \left(\frac1{(an+1)^2} + \frac1{(an+2)^2} + \cdots + \frac1{(an+n)^2}\right)} = \frac43\times54 = 72[/tex]
We have a natural choice of lower and upper bounds for the sum in the denominator:
[tex]\displaystyle \frac1{(an+n)^2} + \cdots + \frac1{(an+n)^2} \\\\ ~ ~ ~ ~ \le \frac1{(an+1)^2} + \cdots + \frac1{(an+n)^2} \\\\ ~ ~ ~ ~ ~ ~ ~ ~ \le \frac1{(an+1)^2} + \cdots + \frac1{(an+1)^2}[/tex]
[tex]\implies \displaystyle \frac{n}{(an+n)^2} \le \frac1{(an+1)^2} + \cdots + \frac1{(an+n)^2} \le \frac{n}{(an+1)^2}[/tex]
and
[tex]\displaystyle \lim_{n\to\infty} n\times\frac{n}{(an+n)^2} = \frac1{(a+1)^2}[/tex]
[tex]\displaystyle \lim_{n\to\infty} n\times\frac{n}{(an+1)^2} = \frac1{a^2}[/tex]
so that by the squeeze/sandwich theorem,
[tex]\displaystyle \frac1{(a+1)^2} \le \lim_{n\to\infty} n \left(\frac1{(an+1)^2} + \frac1{(an+2)^2} + \cdots + \frac1{(an+n)^2}\right) \le \frac1{a^2}[/tex]
[tex]\implies \displaystyle a^2 \le \lim_{n\to\infty} \frac1{n \left(\frac1{(an+1)^2} + \frac1{(an+2)^2} + \cdots + \frac1{(an+n)^2}\right)} \le (a+1)^2[/tex]
and if the middle limit is supposed to evaluate to 72, solving the inequality for a puts it in the interval [6√2 - 1, 6√2] ≈ [7.48528, 8.48528].
Checking against a computer, the solution appears to be a = 8, which agrees with the analysis above. Just not sure how to bridge the gap yet...
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