<s>[INST] Calculate $ \biggr\lfloor \frac{1}{4^{\frac{1}{3}}} + \frac{1}{5^{\frac{1}{3}}} + ... + \frac{1}{1000000^{\frac{1}{3}}} \biggr\rfloor$ 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.
 [/INST] 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.
</s>