Great question on greatest integers

Evaluate $\displaystyle \left \lfloor \sum_{n=1}^{80} \dfrac1{\sqrt n} \right \rfloor$ .

Notation: $\lfloor \cdot \rfloor$ denotes the floor function.

#Calculus

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Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Let's look at $\sum_{n=2}^{80}\frac{1}{\sqrt{n}}$ . It is the same as $\int_2^{80}\frac{1}{\sqrt{\left\lfloor x \right\rfloor}}dx$ .

But $\frac{1}{\sqrt{x}}<\frac{1}{\sqrt{\left\lfloor x \right\rfloor}}<\frac{1}{\sqrt{x-1}}$

So $\int_2^{80}\frac{1}{\sqrt{x}}dx<\int_2^{80}\frac{1}{\sqrt{\left\lfloor x \right\rfloor}}dx<\int_2^{80}\frac{1}{\sqrt{x-1}}dx$

Which easily evaluates to

$15.06<\int_2^{80}\frac{1}{\sqrt{\left\lfloor x \right\rfloor}}dx<15.78$

So $15.06<\sum_{n=2}^{80}\frac{1}{\sqrt{n}}<15.78$

Adding $1$ , we have

$16.06<\sum_{n=1}^{80}\frac{1}{\sqrt{n}}<16.78$

And so $\left\lfloor \sum_{n=1}^{80}\frac{1}{\sqrt{n}} \right\rfloor = 16$

Pedro Cardoso - 3 years, 1 month ago

Thank you sir for so much descriptive answers. Regards, Shashwat

Shashwat Gokhe - 3 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$