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#### mathmaniac

##### Well-known member

- Mar 4, 2013

- 188

sigma(1/n)

Is there a formula for it?

Is there a formula for it?

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- Thread starter
- #1

- Mar 4, 2013

- 188

sigma(1/n)

Is there a formula for it?

Is there a formula for it?

- Mar 31, 2013

- 1,356

There is no closed formula. but approximatelyIs there a formula for it?

gamma + ln (n) when n is too large and gamma is the eulers constant

as Euler?Mascheroni constant - Wikipedia, the free encyclopedia

Also note that the infinite series is divergent, and so that can not possibly have a closed form.sigma(1/n)

Is there a formula for it?

- Mar 31, 2013

- 1,356

The above statement is not quite correct asAlso note that the infinite series is divergent, and so that can not possibly have a closed form.

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.The above statement is not quite correct as

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

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- Mar 4, 2013

- 188

Why not a formula f(n) such that f(n)-f(n-1)=1/n

Why isn't it possible?

Why isn't it possible?

- Mar 31, 2013

- 1,356

I am sorry about my statement. I I meant closed form for the finite sum and then as n tends to infinite. My due apologiesThe FINITE series has a closed form. The INFINITE series does not. There is nothing wrong with what I said.

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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.Why not a formula f(n) such that f(n)-f(n-1)=1/n

Why isn't it possible?

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.

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- Mar 4, 2013

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

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- #11

- Mar 4, 2013

- 188

I think most series including reciprocals have no formulae,but some have and is it possible to check?

- Feb 13, 2012

- 1,704

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

$\chi$ $\sigma$