Find $a$ such that $\lim\limits_{n\to\infty}\frac{\sum\limits_{k=1}^n\sqrt[3]{k}}{n^{\frac73}\sum_{k=1}^n\frac1{(an+k)^2}}=54$

Question

Determine the value of $a > 1$ such that $$\lim_{n\to\infty} \frac{1 + \sqrt[3]{2} + \sqrt[3]{3} + \cdots + \sqrt[3]{n}}{n^{\frac73} \left(\frac1{(an+1)^2} + \frac1{(an+2)^2} + \frac1{(an+3)^2} + \cdots + \frac1{(an+n)^2}\right)} = 54$$

Limits involving sums like these make me think of the Stolz-Cesàro theorem, and I've successfully used S-C (with $a_n$ the sum of cube roots in the numerator and $b_n = n^{\frac43}$) to show that

$$\lim_{n\to\infty} \frac{1 + \sqrt[3]{2} + \sqrt[3]{3} + \cdots + \sqrt[3]{n}}{n^{\frac43}} = \frac34$$

This leaves me with the task of showing

$$\lim_{n\to\infty} \frac{1}{n \left(\frac1{(an+1)^2} + \frac1{(an+2)^2} + \frac1{(an+3)^2} + \cdots + \frac1{(an+n)^2}\right)} = \frac34 \times 54 = 72$$

I was thinking about how to apply S-C again, originally with the choice of $a_n=\frac1{\frac1{(an+1)^2}+\cdots+\frac1{(an+n)^2}}$ and $b_n=n$, but trying to reduce $a_{n+1}-a_n$ looks very tricky.

Instead, I found upper and lower bounds for $a_n$ and using the sandwich theorem, I arrived at

$$\frac n{(an+n)^2} \le \frac1{a_n} \le \frac n{(an+1)^2} \implies a^2 \le \lim_{n\to\infty} \frac{a_n}n \le (a+1)^2$$

The limit must be $72$, which places $a$ in the interval $[6\sqrt2-1,6\sqrt2]\approx[7.4,8.4]$.

The solution given to me is $a=8$ (no explanation included, unfortunately). How do I arrive at this?

Answer 1

Considering $$A_n=\frac{\sum\limits_{k=1}^n\sqrt[3]{k}}{n^{\frac 73}\sum_{k=1}^n\frac1{(an+k)^2}}$$ $$\sum\limits_{k=1}^n\sqrt[3]{k}=H_n^{\left(-\frac{1}{3}\right)}$$ $$\sum_{k=1}^n\frac1{(an+k)^2}=\psi ^{(1)}(a n+1)-\psi ^{(1)}(a n+n+1)$$ $$A_n=\frac{H_n^{\left(-\frac{1}{3}\right)}}{n^{\frac 73} \big[\psi ^{(1)}(a n+1)-\psi ^{(1)}(a n+n+1)\big]}$$ Now, using asymptotics for large values of $n$ (remembering that $a>1$ $$H_p^{\left(-\frac{1}{3}\right)}=\frac{3 p^{4/3}}{4}+\frac{p^{1/3}}{2}+\zeta \left(-\frac{1}{3}\right) +O\left(\frac{1}{p^{2/3}}\right)$$ $$\psi ^{(1)}(p)=\frac{1}{p}+\frac{1}{2 p^2}+O\left(\frac{1}{p^3}\right)$$ Apply the last twice and continue with Taylor series to make $$n^{\frac 73}\sum_{k=1}^n\frac1{(an+k)^2}=\frac{n^{4/3}}{a^2+a}+\frac{1}{2} \left(\frac{1}{(a+1)^2}-\frac{1}{a^2}\right) n^{1/3}+O\left(\frac{1}{n^{5/3}}\right)$$ Now, it becomes simple and you have not only the limit but also the asymptotics.

Answer 2

After consulting similar questions about limits on MSE, I can evidently relate the remaining limit to a definite integral, namely

$$\lim_{n\to\infty} n \sum_{k=1}^n \frac{1}{(an+k)^2} = \lim_{n\to\infty} \frac1n \sum_{k=1}^n \frac1{\left(a+\frac kn\right)^2} = \int_0^1 \frac{dx}{(a+x)^2} = \frac1{a^2+a}$$

It follows that

$$a^2+a=72 \implies a^2 + a - 72 = (a + 9) (a - 8) = 0 \implies \boxed{a=8}$$