Don't Let My Floors Diverge!

Calculus Level 4

n = 1 n n 4 n 8 n 16 n 512 n 1024 m \large \displaystyle \sum_{n=1}^{\infty} \frac{\lfloor \sqrt{n} \rfloor \lfloor \sqrt[4]{n} \rfloor \lfloor \sqrt[8]{n} \rfloor \lfloor \sqrt[16]{n} \rfloor \dots \lfloor \sqrt[512]{n} \rfloor }{\lfloor \sqrt[1024]{n} \rfloor ^m }

Find the minimum integral value of m m such that the above sum converges.

Details and assumptions :

x \bullet \lfloor x \rfloor denotes the floor function.


The answer is 2047.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Abdeslem Smahi
Jul 11, 2015

A really nice approach! That becomes a general fact now.

Kartik Sharma - 5 years, 11 months ago

Well done my friend!

Great, clean, and brief approach! :) :)

Hasan Kassim - 5 years, 11 months ago

Log in to reply

thank you ^^

Abdeslem Smahi - 5 years, 11 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...