harmonic series question

[Kharakov]

One of the standard proofs of the divergence of the harmonic series is:

\(\sum_{1\to \infty} \frac {1}{x} =1 + \frac {1}{2} + \frac {1}{3} + \frac {1}{4} ...= \infty\)

Because it's greater than

\( 1 + \frac {1}{2} + \frac {1}{4} + \frac {1}{4} + \frac {1}{8} + \frac {1}{8}+ \frac {1}{8} + \frac {1}{8} ...\)

Which is 1/2+1/2....

It's also greater than
\( 1 + \frac {1}{3} + \frac{1}{3} + \frac {1}{9} + \frac {1}{9} + \frac {1}{9} + \frac {1}{9} +\frac {1}{9} + \frac {1}{9} + ...\)

which is 2/3 + 2/3 + 2/3....

etc...

The "residuals" (for lack of a better (correct) term) are what is left over when these series are subtracted from the harmonic series. For the first example you get:
r_2=
\(1 + \frac {1}{2} + \frac {1}{3} + \frac {1}{4}+ ...\)
-
\( 1 + \frac {1}{2} + \frac {1}{4} + \frac {1}{4} + ...\)
=

\( r_2 = \frac {1}{12} + \frac {3}{40} + \frac {2}{48} +\frac{1}{56} ...\)

r_3=
\(1 + \frac {1}{2} + \frac {1}{3} + \frac {1}{4}+ ...\)
-
\( 1 + \frac {1}{3} + \frac{1}{3} + \frac {1}{9} + \frac {1}{9} + \frac {1}{9} + \frac {1}{9} +\frac {1}{9} + \frac {1}{9} + ...\)

\(r_3= \frac{1}{6} + \frac{5}{36} +\frac{4}{45} + \frac {3}{54} +...\)


The residuals (r_n) are also divergent.

Is r_n - r_(n+1) divergent?

What is the closed form for r_n - r_(n+a)?

 

[beero1000]

One of the standard proofs of the divergence of the harmonic series is:

\(\sum_{1\to \infty} \frac {1}{x} =1 + \frac {1}{2} + \frac {1}{3} + \frac {1}{4} ...= \infty\)

Because it's greater than

\( 1 + \frac {1}{2} + \frac {1}{4} + \frac {1}{4} + \frac {1}{8} + \frac {1}{8}+ \frac {1}{8} + \frac {1}{8} ...\)

Which is 1/2+1/2....

It's also greater than
\( 1 + \frac {1}{3} + \frac{1}{3} + \frac {1}{9} + \frac {1}{9} + \frac {1}{9} + \frac {1}{9} +\frac {1}{9} + \frac {1}{9} + ...\)

which is 2/3 + 2/3 + 2/3....

etc...

The "residuals" (for lack of a better (correct) term) are what is left over when these series are subtracted from the harmonic series. For the first example you get:
r_2=
\(1 + \frac {1}{2} + \frac {1}{3} + \frac {1}{4}+ ...\)
-
\( 1 + \frac {1}{2} + \frac {1}{4} + \frac {1}{4} + ...\)
=

\( r_2 = \frac {1}{12} + \frac {3}{40} + \frac {2}{48} +\frac{1}{56} ...\)

r_3=
\(1 + \frac {1}{2} + \frac {1}{3} + \frac {1}{4}+ ...\)
-
\( 1 + \frac {1}{3} + \frac{1}{3} + \frac {1}{9} + \frac {1}{9} + \frac {1}{9} + \frac {1}{9} +\frac {1}{9} + \frac {1}{9} + ...\)

\(r_3= \frac{1}{6} + \frac{5}{36} +\frac{4}{45} + \frac {3}{54} +...\)


The residuals (r_n) are also divergent.

Is r_n - r_(n+1) divergent?

What is the closed form for r_n - r_(n+a)?

For large n, \(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots + \frac{1}{n} \sim \ln(n)\) and \(1 + \frac{1}{k} + \dots + \frac{1}{k} + \frac{1}{k^2} + \dots + \frac{1}{k^2} + \dots \sim \frac{k-1}{k}\log_k (n)\), at least up to a small error term each.

Subtracting to get your residuals and then subtracting again just cancels the harmonic terms. The difference of the logs still diverges.

 

[Kharakov]

So nothing Reimanny happens?? What if the residuals have the same number of terms (which means that the harmonics are not aligned)?

I'm going to code it and check it out later.

 

[beero1000]

So nothing Reimanny happens?? What if the residuals have the same number of terms (which means that the harmonics are not aligned)?

I'm going to code it and check it out later.

I don't think so. Of course, we're playing a little fast and loose with rigor here, so to really be sure you'd have to nail down your precise operations/definitions.

 

[Kharakov]

\(1 + \frac{1}{k} + \dots + \frac{1}{k} + \frac{1}{k^2} + \dots + \frac{1}{k^2} + \dots \sim \frac{k-1}{k}\log_k (n)\)

The series increases by (k-1)/k every k^a-1 terms (n= k^a-1). For a large number of terms...

We end up with log_k * (k-1)/k....

With the removed terms we end up having log_k - log_k(n-log_k) difference in number of terms for different k... and as n--> \infty the difference --> 0.

 

[beero1000]

\(1 + \frac{1}{k} + \dots + \frac{1}{k} + \frac{1}{k^2} + \dots + \frac{1}{k^2} + \dots \sim \frac{k-1}{k}\log_k (n)\)

The series increases by (k-1)/k every k^a-1 terms (n= k^a-1). For a large number of terms...

We end up with log_k * (k-1)/k....

With the removed terms we end up having log_k - log_k(n-log_k) difference in number of terms for different k... and as n--> \infty the difference --> 0.

The exact value for \( k^m < n \leq k^{m+1}\) is \(1 + \frac{(k-1)m}{k} + \frac{n - k^m}{k^{m+1}}\).

In terms of just n and k, you'd get \(\frac{k-1}{k}( \lfloor \log_k(n-1) \rfloor + 1) + \frac{n}{k^{\lfloor \log_k(n-1)\rfloor+1}} \) so the only term that really matters is the log term and the differences still diverge.