Calculate $ \biggr\lfloor \frac{1}{4^{\frac{1}{3}}} + \frac{1}{5^{\frac{1}{3}}} + \frac{1}{6^{\frac{1}{3}}} + ... + \frac{1}{1000000^{\frac{1}{3}}} \biggr\rfloor$
I am just clueless. I just have some random thoughts. I can find the sum of this series and then put in the value $1000000$ and then find the floor of that. To do this, I would have to telescope this series which seems impossible to me. But is there any other way to directly find the floor without finding the general sum?
I am very new to calculus, so please provide hints and answers that dio not involve calculus.
Since $(x^a)' = ax^{a-1}$, $a\int_n^{n+1} x^{a-1}dx =(n+1)^a-n^a $.
Therefore
$\begin{array}\\ v^a-u^a &=\sum_{n=u}^{v-1}((n+1)^a-n^a)\\ &=\sum_{n=u}^{v-1}a\int_n^{n+1} x^{a-1}dx\\ \end{array} $
If $a > 1$, since $x^{a-1}$ is increasing, $n^{a-1} \lt \int_n^{n+1} x^{a-1}dx \lt (n+1)^{a-1} $.
If $a < 1$, since $x^{a-1}$ is decreasing, $n^{a-1} \gt \int_n^{n+1} x^{a-1}dx \gt (n+1)^{a-1} $.
In this case, $a-1 = -1/3 < 0$. Therefore $v^a-u^a =\sum_{n=u}^{v-1}a\int_n^{n+1} x^{a-1}dx \lt \sum_{n=u}^{v-1}an^{a-1} $ so $\sum_{n=u}^{v-1}n^{a-1} \gt \frac1{a}(v^a-u^a) $.
Similarly, $v^a-u^a =\sum_{n=u}^{v-1}a\int_n^{n+1} x^{a-1}dx \gt \sum_{n=u}^{v-1}a(n+1)^{a-1} = a\sum_{n=u+1}^{v}n^{a-1} = a\sum_{n=u}^{v-1}n^{a-1}-a(u^{a-1}-v^{a-1}) $ so $\sum_{n=u}^{v-1}n^{a-1} \lt \frac1{a}((v-1)^a-u^a)+(u^{a-1}-v^{a-1}) $.
The reverse inequalities hold if $a > 1$.
In this case, $a-1=-1/3, a = 2/3, v=10^6+1, u=4$, so, if $s = \sum_{n=u}^{v-1}n^{a-1}$, then $s \gt \frac1{2/3}((10^6)^{2/3}-4^{2/3}) \gt \frac32(10^4-2.519...) \approx 14996.220 $ and $s \lt \frac1{2/3}((10^6)^{2/3}-4^{2/3})+(4^{-1/3}-(10^6)^{-1/3}) \approx 14996.220+0.619 = 14996.839 $
so $\lfloor s \rfloor =14996 $.
If the lower and upper bounds surrounded an integer, I would manually compute the first few terms until the bounds for the remaining terms are close enough that the floor is determined.