Is There a Formula for Summing 1/n Using the Digamma Function?

  • Context: MHB 
  • Thread starter Thread starter mathmaniac1
  • Start date Start date
  • Tags Tags
    Formula Summation
Click For Summary

Discussion Overview

The discussion revolves around the question of whether there is a formula for summing the series 1/n, particularly in relation to the digamma function. Participants explore the nature of the series, its convergence, and potential closed forms, addressing both finite and infinite cases.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants assert that there is no closed formula for the infinite series of 1/n, noting its divergence.
  • Others argue that while the infinite series diverges, the finite series does have a closed form, specifically mentioning the formula for the sum of the first n integers.
  • A participant questions the possibility of finding a function f(n) such that f(n) - f(n-1) = 1/n, suggesting that it may not be algebraically feasible.
  • Another participant introduces the concept of the nth Harmonic number, H_n, as a representation of the finite sum of the series.
  • One post references a source that claims a relationship involving the digamma function and the sum of the series, indicating a potential connection worth exploring.

Areas of Agreement / Disagreement

Participants express differing views on the existence of a closed formula for the series. While there is agreement that the infinite series diverges, there is contention regarding the implications of this divergence and the nature of closed forms for finite sums.

Contextual Notes

Some statements regarding the divergence of series and the existence of closed forms depend on specific definitions and interpretations, which are not fully resolved in the discussion.

mathmaniac1
Messages
158
Reaction score
0
sigma(1/n)

Is there a formula for it?
 
Physics news on Phys.org
mathmaniac said:
sigma(1/n)

Is there a formula for it?

Also note that the infinite series is divergent, and so that can not possibly have a closed form.
 
Prove It said:
Also note that the infinite series is divergent, and so that can not possibly have a closed form.

The above statement is not quite correct as

sigma n = n(n+1)/2 is divergergent but it has a colsed form
 
kaliprasad said:
The above statement is not quite correct as

sigma n = n(n+1)/2 is divergergent but it has a colsed form

The FINITE series has a closed form. The INFINITE series does not. There is nothing wrong with what I said.
 
Why not a formula f(n) such that f(n)-f(n-1)=1/n
Why isn't it possible?
 
Prove It said:
The FINITE series has a closed form. The INFINITE series does not. There is nothing wrong with what I said.

I am sorry about my statement. I I meant closed form for the finite sum and then as n tends to infinite. My due apologies
 
mathmaniac said:
Why not a formula f(n) such that f(n)-f(n-1)=1/n
Why isn't it possible?

It is neither algebraically possible to obtain a homogeneous difference equation by symbolic differencing, nor to find an elementary particular solution to attempt the method of undetermined coefficients.

So what we do is write:

$$\sum_{k=1}^n\frac{1}{k}=H_n$$

where $H_n$ is the $n$th Harmonic number - Wikipedia, the free encyclopedia.
 
Is it possible to figure out whether an inductive formula exists for sigma something?
 
  • #10
Do you find a pattern from which you can infer an induction hypothesis?
 
  • #11
Looking for a pattern is not easy,how do you know when to stop looking and conclude there is no formula?
I think most series including reciprocals have no formulae,but some have and is it possible to check?
 
  • #12
In...

http://mathhelpboards.com/discrete-mathematics-set-theory-logic-15/difference-equation-tutorial-draft-part-i-426.html#post2494

... it has been demonstrated that is...

$\displaystyle \sum_{k=1}^{n} \frac{1}{k} = \phi (n) + \gamma\ (1)$

... where $\phi(*)$ is the digamma function, defined as...

$\displaystyle \phi(x) = \frac{d}{d x} \ln x!\ (2)$

... being...

$\displaystyle x! = \int_{0}^{\infty} t^{x}\ e^{- t}\ dt\ (3)$

Kind regards

$\chi$ $\sigma$
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 15 ·
Replies
15
Views
3K
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K