Greatest integer less than or equal to $\sum_{n=1}^{9999}\frac{1}{n^{1/4}}$ This is a PhD entrance question of TIFR 2020. The question requires the explicit answer. I know that the partial sums are evaluated using Abel's formula in Number Theory but I believe there may be better methods for series of this form. Can anyone share their thoughts?
 A: Hint: Consider the function $f(x):=\frac43\cdot x^{\frac34}$ and use the Mean Value Theorem in order to deduce that $$\frac{1}{\sqrt[4]{r+1}}=f'(r+1)<\frac{f(r+1)-f(r)}{r+1-r}<f'(r)=\frac1{\sqrt[4]{r}}\iff\fbox{$\displaystyle \frac{1}{\sqrt[4]{r+1}}< f(r+1)-f(r)<\frac1{\sqrt[4]{r}}$}$$ You can now sum and use the fact that almost everything will telescope.
A: Compare the sum with the appropriate definite integrals:
$$\sum_{n=1}^{9999}\frac{1}{n^{1/4}}>\int_1^{10000}\frac{dx}{x^{1/4}}=\frac{4}{3}x^{3/4}\bigg|_1^{10000}=\frac{4}{3}\cdot 999=1332$$
Also:
$$\begin{align}
\sum_{n=1}^{9999}\frac{1}{n^{1/4}}
&< \sum_{n=1}^{10000}\frac{1}{n^{1/4}} \\
&= 1+\sum_{n=2}^{10000}\frac{1}{n^{1/4}} \\
&< 1+\int_1^{10000}\frac{dx}{x^{1/4}} \\
&= 1+\frac{4}{3}x^{3/4}\bigg|_1^{10000} \\
&= 1+\frac{4}{3}\cdot 999 \\
&=1333
\end{align}$$
So, the sum is between $1332$ and $1333$ and so its integral part is $1332$.
A: Here's another way to consider stinking Bishops answer.  This is a derivative answer and the exact same as Stinking Bishop's.  I'm just squinting and looking at it from a different angle.

$c_1=\frac 1{(n+1)^{\frac 14}} \le \frac 1{x^{\frac 14}} \le \frac 1{n^{\frac 14}}=c_2$


$c_1 \le \inf_{x\in [n,n+1]}\frac 1{x^{\frac 14}} \le \sup_{x\in [n,n+1]}\frac 1{x^{\frac 14}} \le c_2$


$\int_{n}^{n+1} c_1dx \le \int_{n}^{n+1}\frac 1{x^{\frac 14}}dx \le \int_n^{n+1} c_2 dx$

Now $\int_a^b C dx = C[b-a]$ so $\int_{n}^{n+1} c_1dx=c_1= \frac 1{(n+1)^{\frac 14}}$ and  $\int_n^{n+1} c_2 dx=\frac 1{n^{\frac 14}}$ so

$\frac 1{(n+1)^{\frac 14}}= \int_{n}^{n+1}\frac 1{x^{\frac 14}}dx \le \frac 1{n^{\frac 14}}$


$\sum\limits_{n=1}^{9999}\frac 1{(n+1)^{\frac 14}}\le \sum\limits_{n=1}^{9999} \int_{n}^{n+1}\frac 1{x^{\frac 14}}dx=\int_1^{10000}\frac 1{x^{\frac 14}} dx\le  \sum\limits_{n=1}^{9999}\frac 1{n^{\frac 14}}$

As noted $\int_1^{10000}\frac 1{x^{\frac 14}} dx= 1332$
But also note
$\sum\limits_{n=1}^{9999}\frac 1{(n+1)^{\frac 14}}$ may be reindexed as $\sum\limits_{n=2}^{10000}\frac 1{n^{\frac 14}}$ which is equal to $\sum\limits_{n=1}^{9999}\frac 1{n^{\frac 14}} + \frac 1{10000^{\frac 14}} - \frac 1{1^{\frac 14}}= \sum\limits_{n=1}^{9999}\frac 1{n^{\frac 14}}- 0.9$.
So we have

$\sum\limits_{n=1}^{9999}\frac 1{n^{\frac 14}}- 0.9\le 1332 \le \sum\limits_{n=1}^{9999}\frac 1{n^{\frac 14}}$

And it is easily verified that if $M - 1< M-0.9 \le n \le M$ then $M< n+1 \le M+1$ and so $n\le M< n+1$ so $\lfloor M\rfloor=n$.

So $\lfloor \sum\limits_{n=1}^{9999}\frac 1{n^{\frac 14}}\rfloor =1332$.

