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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.

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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 $n\to\infty$, $$\frac {\left(\sum_{r=1}^n r^4\right)\cdot\left(\sum_{r=1}^n r^5\right)}{\left(\sum_{r=1}^n r^t\right)\cdot\left(\sum_{r=1}^n r^{9-t}\right)} =\frac{\frac{1}{n^5}\left(\sum_{r=1}^n r^4\right)\cdot\frac{1}{n^6}\left(\sum_{r=1}^n r^5\right)}{\frac{1}{n^{t+1}}\left(\sum_{r=1}^n r^t\right)\cdot\frac{1}{n^{10-t}}\left(\sum_{r=1}^n r^{9-t}\right)} \to\frac{F(4)\cdot F(5)}{F(t)\cdot F(9-t)}.$$

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  • $\begingroup$ Oh yes I should have thought of Riemann sums and integrals $\endgroup$ – Rohan Shinde Jul 20 '18 at 15:53
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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.

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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 $t$.

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