Solution to integer part of fourth root reciprocals

This is a solution to the puzzle posted earlier.

The problem:

What is the integer part of
$$1 + \frac{1}{\sqrt[4]{2}} + \frac{1}{\sqrt[4]{3}} + \dots + \frac{1}{\sqrt[4]{10000}}$$
i.e. the integer part of
$$\sum_{k=1}^{10000} \frac{1}{\sqrt[4]{k}}$$

Solution

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The typical way to approach this problem is to approximate the sum by an integral and find appropriate upper and lower bounds.

We will essentially do the same, but presented differently.

Consider the function $\displaystyle f(x) = \frac{4x^{3/4}}{3}$. The derivative of $f$ is $\displaystyle f'(x) = \frac{1}{\sqrt[4]{x}}$

Now using the mean value theorem, and the fact that $\frac{1}{\sqrt[4]{x}}$ is decreasing, we have the following:
$$\frac{1}{\sqrt[4]{k+1}} \lt f(k+1) - f(k) \lt \frac{1}{\sqrt[4]{k}} \quad \quad (1)$$
Using the left side of the inequality, and adding from $k=1$ to $9999$ we get
$$\sum_{k=2}^{10000} \frac{1}{\sqrt[4]{k}} \lt f(10000) - f(1) = \frac{4}{3}(10^3 - 1) = 1332$$
Thus adding $1$ to both sides,
$$\sum_{k=1}^{10000} \frac{1}{\sqrt[4]{k}} \lt 1333$$
Using the right side of $(1)$, and adding from $k=1$ to $9999$ we get

$$\sum_{k=1}^{9999} \frac{1}{\sqrt[4]{k}} \gt f(10000) - f(1) = 1332$$
Thus adding $\frac{1}{\sqrt[4]{10000}} = \frac{1}{10}$ to both sides,
$$\sum_{k=1}^{10000} \frac{1}{\sqrt[4]{k}} \gt 1332 + \frac{1}{10}$$
Thus the integer part is $1332$.