Question

`\rm For \: a \in\R |a| > 1, let \\ \rm \lim_(n \to \infty ) \left \lgroup \rm\frac{1 + \sqrt[3]{2} + \dots + \sqrt[3]{n} }{ {n}^{ (7)/(3) } \left( \frac{1}{(an + 1 {)}^(2) } + \frac{1}{(an + 2 {)}^(2) } + \dots \frac{1}{(an + n{)}^(2) } \right) } \right\rgroup = 54 \\ \rm then \: the \: possible \: value \: of \: a \: is`​

Answer 1

A possible starting point:

Split up the limit as

`\displaystyle \lim_(n\to\infty) \frac{1 + \sqrt[3]{2} + \cdots + \sqrt[3]{n}}{n^(4/3)} * \lim_(n\to\infty) \frac1{n \left(\frac1{(an+1)^2} + \frac1{(an+2)^2} + \cdots + \frac1{(an+n)^2}\right)} = 54`

Consider the first limit,

`\displaystyle \lim_(n\to\infty) \frac{1 + \sqrt[3]{2} + \cdots + \sqrt[3]{n}}{n^(4/3)}`

Refer to the Stolz-Cesàro theorem, which says

`\displaystyle \lim_(n\to\infty) (a_n)/(b_n) = \lim_(n\to\infty)(a_(n+1)-a_n)/(b_(n+1)-b_n)`

where
`a_n` and
`b_n` are two real sequences, with
`b_n` monotone and divergent. In this case,

`a_n = 1+\sqrt[3]{2}+\sqrt[3]{3}+\cdots+\sqrt[3]{n}`

`b_n = n^(4/3)`

Applying S-C, we get

`\displaystyle \lim_(n\to\infty) \frac{\sqrt[3]{n+1}}{(n+1)^(4/3) - n^(4/3)} = \lim_(n\to\infty) ((n+1)^(1/3))/((n+1)^(4/3) - n^(4/3))`

Recalling the difference of cubes identity,

`a^3 - b^3 = (a - b) (a^2 + ab + b^2)`

we can rewrite the limit as

`\displaystyle \lim_(n\to\infty) ((n+1)^3 + (n+1)^(5/3) n^(4/3) + (n+1)^(1/3) n^(8/3))/((n+1)^4 - n^4)`

and dividing uniformly through the limand by (n + 1)³ yields

`\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) - (n^4)/((n+1)^3)}`

Now,

`n^4 = (n+1)^4 - 4n^3 - 6n^2 - 4n - 1`

`\implies (n^4)/((n+1)^3) = (n+1) - (4n^3+6n^2+4n+1)/((n+1)^3)`

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

`\displaystyle \lim_(n\to\infty) \frac{1 + \sqrt[3]{2} + \cdots + \sqrt[3]{n}}{n^(4/3)} = \frac34`

It remains to determine the value of a such that

`\displaystyle \lim_(n\to\infty) \frac1{n \left(\frac1{(an+1)^2} + \frac1{(an+2)^2} + \cdots + \frac1{(an+n)^2}\right)} = \frac43*54 = 72`

We have a natural choice of lower and upper bounds for the sum in the denominator:

`\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}`

`\implies \displaystyle (n)/((an+n)^2) \le \frac1{(an+1)^2} + \cdots + \frac1{(an+n)^2} \le (n)/((an+1)^2)`

and

`\displaystyle \lim_(n\to\infty) n*(n)/((an+n)^2) = \frac1{(a+1)^2}`

`\displaystyle \lim_(n\to\infty) n*(n)/((an+1)^2) = \frac1{a^2}`

so that by the squeeze/sandwich theorem,

`\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}`

`\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`

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...