Find $t$ such that $\lim_{n\to\infty} \frac {\sum_{r=1}^n r^4\cdot\sum_{r=1}^n r^5}{\sum_{r=1}^n r^t\cdot\sum_{r=1}^n r^{9-t}}=\frac 45$

Question

Find $t$ such that $$\lim_{n\to\infty} \frac {\left(\sum_{r=1}^n r^4\right)\cdot\left(\sum_{r=1}^n r^5\right)}{\left(\sum_{r=1}^n r^t\right)\cdot\left(\sum_{r=1}^n r^{9-t}\right)}=\frac 45.$$

At first sight this question scared the hell out of me. I tried using the general known formulas like $$\sum_{r=1}^n r^4=\frac {n(n+1)(2n+1)(3n^2+3n-1)}{6}$$ and $$\sum_{r=1}^n r^5=\frac {n^2(n+1)^2(2n^2+2n-1)}{12}.$$

But the denominator portion really doesn't go with it. I tried to write it in form of integrals. I also searched the internet for some information but it dealt higher level calculus relating the harmonic functions, Bernoulli numbers and the Riemann zeta function. I read about it but couldn't get much out of it.

Any help would be greatly appreciated. Thanks.

You probably just need to prove $\sum_{k=1}^n k^r \sim c_r n^{r+1}$ for some $c_r$ that does not depend on $n$. — Arin Chaudhuri, Jul 20 '18 at 15:43
Use https://math.stackexchange.com/questions/469885/the-limit-of-a-sum-sum-k-1n-fracnn2k2/469886#469886 — lab bhattacharjee, Jul 20 '18 at 15:44

Robert Z · Accepted Answer · 2018-07-20T15:57:02.070

Hint. Note that for $a>0$ $$F(a):=\lim_{n\to\infty} \frac{1}{n^{a+1}}\left(\sum_{r=1}^n r^a\right)= \lim_{n\to\infty} \frac{1}{n}\left(\sum_{r=1}^n \left(\frac{r}{n}\right)^a\right)\to \int_0^1x^a dx=\frac{1}{a+1}.$$ Then, as $n\to\infty$, $$\frac {\left(\sum_{r=1}^n r^4\right)\cdot\left(\sum_{r=1}^n r^5\right)}{\left(\sum_{r=1}^n r^t\right)\cdot\left(\sum_{r=1}^n r^{9-t}\right)} =\frac{\frac{1}{n^5}\left(\sum_{r=1}^n r^4\right)\cdot\frac{1}{n^6}\left(\sum_{r=1}^n r^5\right)}{\frac{1}{n^{t+1}}\left(\sum_{r=1}^n r^t\right)\cdot\frac{1}{n^{10-t}}\left(\sum_{r=1}^n r^{9-t}\right)} \to\frac{F(4)\cdot F(5)}{F(t)\cdot F(9-t)}.$$

Oh yes I should have thought of Riemann sums and integrals – Rohan Shinde Jul 20 '18 at 15:53 — Rohan Shinde, Jul 20 '18 at 15:53

score 2 · Answer 2 · answered Jul 20 '18 at 15:49

The skeleton:

First thought is to apply integrals:

$$ \dfrac{\int\limits_1^n x^4 dx \int\limits_1^n x^5 dx}{\int\limits_1^n x^t dx \int\limits_1^n x^{9-t} dx} \sim \dfrac{\dfrac{n^5}{5} \cdot \dfrac{n^6}{6}}{\dfrac{n^{t+1}}{t+1} \cdot \dfrac{n^{9-t+1}}{9-t+1}} $$

Then we'll have fraction $$ \dfrac{(9-t+1)(t+1)}{30}=\dfrac{4}{5} = \dfrac{24}{30} $$

Then quadratic equation $$(9-t+1)(t+1) = 24$$ So $t=2$, $t=7$ should work.

score 0 · Answer 3 · answered Jul 21 '18 at 02:42

The key here is Cesaro-Stolz. Using Cesaro-Stolz we have $$f(t) =\lim_{n\to\infty} \frac{1}{n^{t+1}}\sum_{r=1}^{n}r^t=\lim_{n\to \infty} \frac{n^t} {n^{t+1}-(n-1)^{t+1}}=\frac{1}{t+1}$$ Dividing the numerator and denominator of the given expression by $n^{11}$ we can see that the desired limit is $$\frac{f(4)f(5)}{f(t)f(9-t)}$$ and now you can equate this to $4/5$ and get $t$.

Find $t$ such that $\lim_{n\to\infty} \frac {\sum_{r=1}^n r^4\cdot\sum_{r=1}^n r^5}{\sum_{r=1}^n r^t\cdot\sum_{r=1}^n r^{9-t}}=\frac 45$

3 Answers3