Prove that is convergent.

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- May 22nd 2009, 10:28 PMNonCommAlgInfinite series (4)
Prove that is convergent.

- May 23rd 2009, 10:47 PMsimplependulum
can we use the inequality ( the equality holds only when n = 1 ) ?

the next step we have to go is just considering whether converges - May 23rd 2009, 11:16 PMNonCommAlg
- May 24th 2009, 06:47 AMSampras
Maybe use the integral test and we know that .

- May 25th 2009, 12:47 AMchisigma
If we indicate with...

(1)

... the 'armonic number' of index n, the 'infinite series' is...

(2)

Remembering that the generating function of armonic numbers is...

(3)

... we derive in two steps...

(5)

(6)

Now we use the identity...

(7)

... and (6) to arrive at the result...

(8)

Kind regards

- Aug 6th 2009, 10:26 PMluobo
Investigate the convergence of series .

It is well known that , where is the Euler-Mascheroni constant.

Obviously series converges since , knowing exists.

Now investigate the convergence of series , leaving to check if the following integral exists

. Integral by part, it is easy to know it exists.

Therefore, converges. - Aug 7th 2009, 12:43 AMMoo
Luobo,

What you did is correct.

But it doesn't help for the present problem, since we want to prove if the**series**is convergent, not the sequence .

chisigma, how do you prove that converges ? - Aug 7th 2009, 04:46 AMluobo
Moo,

I indeed is talking about series. For clarity, it is rearranged below:

(1) Series is convergent.

{ Note: }

The first term converges knowing

(2) Series is also convergent.

It converges since is bounded.

(3) Series is convergent

- Aug 7th 2009, 09:33 PMchisigma
... all right!... with little patience [or even using Wolfram Integrator (Wink) ...] you find that...

(1)

... where is the Polylogarithm function of order two, defined in the complex plane as...

(2)

Now the series (2) converges for so that and both exist...

Kind regards

- Aug 8th 2009, 11:05 PMchisigma
Following the same procedure we arrive to demonstrate that is...

(1)

These are interesting results that are not isolated, because the same procedure can be used to arrive to even better results (Wink) ...

Kind regards

- Aug 9th 2009, 12:40 AMchisigma
An example... from the well known expression...

(1)

... we derive...

(2)

Now from (2) and the identity...

(3)

... we derive...

(4)

... and from (2) and the identity...

(5)

... we derive...

(6)

Kind regards

- Aug 9th 2009, 10:42 AMNonCommAlg
- Aug 9th 2009, 05:05 PMluobo
NonConAlg,

You are right. Thanks a lot. Maybe we can switch to Abel's test. Abel's Test. Check below.

(1) Partial sum is bounded.

= Im = Im

(2) Series is convergent.

{ Note: }

The second term converges obviously. The first term also converges according to Abel's test:

(Condition 1) Sequence is decreasing

(Condition 2)

(Condition 3) Partial sum is bounded

(3) Series is also convergent.

(Condition 1) Sequence is decreasing for

(Condition 2)

(Condition 3) Partial sum is bounded

(4) Series is convergent

- Aug 9th 2009, 05:49 PMhalbard
The sequence is monotonically decreasing to .

Let . Then .

Thus for all , i.e. the partial sums of are bounded.

By Dirichlet's test, converges. - Aug 9th 2009, 06:28 PMluobo
halbard,

I was thinking about doing the same thing (Happy). My reference about the Abel's Test from a website appears ought to be the Dirichlet's test.

Luobo