<s>[INST] Integration and summation: prove $399< \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\dots+\frac{1}{\sqrt{40000}}<400$ is false. I need help proving that the following statement is false:
$$
399 < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} 
    + \frac{1}{\sqrt{3}} + \dots + \frac{1}{\sqrt{40000}} < 400.
$$
I tried to bound the summation between two integrals
$$
\text{1)} \int_1^{40001} \frac{2}{\sqrt{x}}
\qquad \text{and} \qquad
\text{2)} \int_0^{40000} \frac{2}{\sqrt{x}}
$$
and have convinced myself that the statement is true, but this is not the correct answer as per my textbook.
 [/INST] Your sum is equal to
$$1 + \sum_{n=2}^{40000} \frac{1}{\sqrt{n}} < 1 + \int_1^{40000} \frac{1}{\sqrt{x}} \; dx = 1+ \left.2\sqrt{x}\right|_{x=1}^{40000} = 1+(2\sqrt{40000} - 2) = 399.$$
</s>