Prove that is convergent.
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
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.
Moo,
I indeed is talking about series. For clarity, it is rearranged below:
(1) Series is convergent.
{ Note: }
The first term converges knowing is bounded, the second term converges even more obviously.
(2) Series is also convergent.
It converges since is bounded.
(3) Series is convergent
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
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