Great question on greatest integers

Evaluate n=1801n \displaystyle \left \lfloor \sum_{n=1}^{80} \dfrac1{\sqrt n} \right \rfloor .


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

#Calculus

Note by Shashwat Gokhe
3 years, 1 month ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Let's look at n=2801n\sum_{n=2}^{80}\frac{1}{\sqrt{n}}. It is the same as 2801xdx\int_2^{80}\frac{1}{\sqrt{\left\lfloor x \right\rfloor}}dx.

But 1x<1x<1x1\frac{1}{\sqrt{x}}<\frac{1}{\sqrt{\left\lfloor x \right\rfloor}}<\frac{1}{\sqrt{x-1}}

So 2801xdx<2801xdx<2801x1dx\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<2801xdx<15.7815.06<\int_2^{80}\frac{1}{\sqrt{\left\lfloor x \right\rfloor}}dx<15.78

So 15.06<n=2801n<15.7815.06<\sum_{n=2}^{80}\frac{1}{\sqrt{n}}<15.78

Adding 11, we have

16.06<n=1801n<16.7816.06<\sum_{n=1}^{80}\frac{1}{\sqrt{n}}<16.78

And so n=1801n=16\left\lfloor \sum_{n=1}^{80}\frac{1}{\sqrt{n}} \right\rfloor = 16

Pedro Cardoso - 3 years, 1 month ago

Log in to reply

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

Shashwat Gokhe - 3 years, 1 month ago
×

Problem Loading...

Note Loading...

Set Loading...