Divisor function (Corrected)

Probability Level 5

If

A = ∑ n = 1 ∞ ( ∑ k = 1 n 2 d ( k ) − 2 M n ( k ) k ) n − 2 A=\sum _{ n=1 }^{ \infty } \left( \sum _{ k=1 }^{ n^{ 2 } } \frac { d\left( k \right) -2M_{n}\left( k \right) }{ k } \right) n^{ -2 }

where d ( k ) d(k) is the number of divisors of k k and M n ( k ) M_{n}(k) is the number of divisors of k k that are greater than n n , find ⌊ 100 A ⌋ \left\lfloor 100A \right\rfloor .

Clue: Understand the derivation of ζ ( s ) 2 = ∑ k = 1 ∞ d ( k ) k s { \zeta (s) }^{ 2 }=\sum _{ k=1 }^{ \infty }{ \frac { d(k) }{ { k }^{ s } } }

I just realised that there was a typo in my previous question, so I'm reposting it. Sorry to those who wasted their time on a flawed problem.


The answer is 459.

Julian Poon by Julian Poon , 20, Singapore

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

Julian Poon
Oct 12, 2015

Let's first consider the derivation of ζ ( s ) 2 = ∑ k = 1 ∞ d ( k ) k s { \zeta (s) }^{ 2 }=\sum _{ k=1 }^{ \infty }{ \frac { d(k) }{ { k }^{ s } } }

In this product ( 1 1 s + 1 2 s + 1 3 s . . . ) ( 1 1 s + 1 2 s + 1 3 s . . . ) \left( \frac { 1 }{ { 1 }^{ s } } +\frac { 1 }{ { 2 }^{ s } } +\frac { 1 }{ { 3 }^{ s } } ... \right) \left( \frac { 1 }{ { 1 }^{ s } } +\frac { 1 }{ { 2 }^{ s } } +\frac { 1 }{ { 3 }^{ s } } ... \right) , how many times would 1 10 s \frac{1}{{10}^{s}} appear upon expansion?

Since 1 1 s ⋅ 1 10 s = 1 2 s ⋅ 1 5 s = 1 10 s \frac { 1 }{ { 1 }^{ s } } \cdot \frac { 1 }{ { 10 }^{ s } } =\frac { 1 }{ { 2 }^{ s } } \cdot \frac { 1 }{ { 5 }^{ s } } =\frac { 1 }{ { 10 }^{ s } } , 1 10 s \frac{1}{{10}^{s}} would appear 4 4 times.

From this, we can see that the number of times that 1 n s \frac{1}{{n}^{s}} would appear in the product ( 1 1 s + 1 2 s + 1 3 s . . . ) ( 1 1 s + 1 2 s + 1 3 s . . . ) \left( \frac { 1 }{ { 1 }^{ s } } +\frac { 1 }{ { 2 }^{ s } } +\frac { 1 }{ { 3 }^{ s } } ... \right) \left( \frac { 1 }{ { 1 }^{ s } } +\frac { 1 }{ { 2 }^{ s } } +\frac { 1 }{ { 3 }^{ s } } ... \right) is equal to the number of divisors n n has, hence ( 1 1 s + 1 2 s + 1 3 s . . . ) ( 1 1 s + 1 2 s + 1 3 s . . . ) = ∑ k = 1 ∞ d ( k ) k s \left( \frac { 1 }{ { 1 }^{ s } } +\frac { 1 }{ { 2 }^{ s } } +\frac { 1 }{ { 3 }^{ s } } ... \right) \left( \frac { 1 }{ { 1 }^{ s } } +\frac { 1 }{ { 2 }^{ s } } +\frac { 1 }{ { 3 }^{ s } } ... \right) =\sum _{ k=1 }^{ \infty }{ \frac { d(k) }{ { k }^{ s } } }

Now what happens if instead of the square of an infinite sum (The zeta function), we have the square of a harmonic number?

H n 2 = ( 1 1 + 1 2 + 1 3 . . . + 1 n ) ( 1 1 + 1 2 + 1 3 . . . + 1 n ) { H }_{ n }^{ 2 }=\left( \frac { 1 }{ { 1 } } +\frac { 1 }{ { 2 } } +\frac { 1 }{ { 3 } } ...+\frac { 1 }{ { n } } \right) \left( \frac { 1 }{ { 1 } } +\frac { 1 }{ { 2 } } +\frac { 1 }{ { 3 } } ...+\frac { 1 }{ { n } } \right)

Upon expansion, if surfaces that H n 2 = ( ∑ k = 1 n 2 d ( k ) k ) − 2 ( ∑ k = 1 + n n 2 M n ( k ) k ) { H }_{ n }^{ 2 }=\left( \sum _{ k=1 }^{ n^{ 2 } } \frac { d\left( k \right) }{ k } \right) -2\left( \sum _{ k=1+n }^{ n^{ 2 } } \frac { M_{ n }\left( k \right) }{ k } \right)

Our sum can thus be rewritten as

A = ∑ n = 1 ∞ ( H n 2 ) n − 2 A=\sum _{ n=1 }^{ \infty } \left( { H }_{ n }^{ 2 } \right) n^{ -2 }

This sum is known to converge to 17 360 π 4 \frac { 17 }{ 360 } { \pi }^{ 4 } , see the derivation in this paper

Hence, ⌊ 100 A ⌋ = 459 \left\lfloor 100A \right\rfloor=\boxed{459}

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...