Let $$M=\left\lfloor \frac{1}{1^{\frac{2}{3}}}+\frac{1}{2^{\frac{2}{3}}}+\frac{1}{3^{\frac{2}{3}}}+...+\frac{1}{1000^{\frac{2}{3}}}\right\rfloor$$.
Find $M$
My Attempt
Let $f(x)=x^{-\frac{2}{3}}$
Since $f''(x)>0$
we have
$\sum_{r=1}^{999}r^{-\frac{2}{3}}>\int_{1}^{999}x^{-\frac{2}{3}}dx$
Also, $\sum_{r=2}^{999}r^{-\frac{2}{3}}<\int_{1}^{1000}x^{-\frac{2}{3}}dx$
I am stuck here. Don't know how to proceed
Let $$S = \sum_{r=1}^{1000} r^{-2/3}.$$ Then because for each positive integer $r$, $$r^{-2/3} = \int_{x=r}^{r+1} \lfloor x \rfloor^{-2/3} \, dx > \int_{x=r}^{r+1} x^{-2/3} \, dx,$$ it follows that $$S > \sum_{r=1}^{1000} \int_{x=r}^{r+1} x^{-2/3} \, dx = \int_{x=1}^{1001} x^{-2/3} \, dx = 3(1001^{1/3} - 1) > 3(1000^{1/3} - 1) = 27.$$ Conversely, $$r^{-2/3} = \int_{x=r-1}^r \lceil x \rceil^{-2/3} \, dx < \int_{x=r-1}^r x^{-2/3} \, dx,$$ hence $$S = 1 + \sum_{r=2}^{1000} r^{-2/3} < 1 + \int_{x=1}^{1000} x^{-2/3} \, dx = 1 + 27 = 28.$$ Note that we must "peel off" the first term in the summation because the upper bound $$1 < \int_{x=0}^1 x^{-2/3} \, dx = 3$$ is too loose. This establishes $$27 < S < 28,$$ hence $M = \lfloor S \rfloor = 27.$