# summation formula

#### mathmaniac

##### Active member
sigma(1/n)

Is there a formula for it?

#### Prove It

##### Well-known member
MHB Math Helper
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.

##### Well-known member
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

#### Prove It

##### Well-known member
MHB Math Helper
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.

#### mathmaniac

##### Active member
Why not a formula f(n) such that f(n)-f(n-1)=1/n
Why isn't it possible?

##### Well-known member
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

#### MarkFL

Staff member
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:

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

where $H_n$ is the $n$th Harmonic number - Wikipedia, the free encyclopedia.

#### mathmaniac

##### Active member
Is it possible to figure out whether an inductive formula exists for sigma something?

#### MarkFL

Staff member
Do you find a pattern from which you can infer an induction hypothesis?

#### mathmaniac

##### Active member
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?

#### chisigma

##### Well-known member
In...

http://mathhelpboards.com/discrete-...ation-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

$\chi$ $\sigma$