From: Helmut Kahovec <helmut.kahovec@teleweb.at>
Subject: Re: Raabe's Theorem
Date: Fri, 28 Jan 2000 11:21:58 GMT
Newsgroups: sci.math
Summary: [missing]

*******************************************************
*                                                     *
* Please, use a fixed font when reading this posting! *
*                                                     *
*******************************************************

TAZ wrote:

>| I posted a question on this site and it was solved using
>| Raabe's Theorem.  I have never studied this theorem,
>| however, it seems very useful.  The problem is that upon
>| searching for it on the web, I found a diiferent form than
>| I read here.
>| 
>| Here:    lim   n[a(n)/a(n+1) -1]
>| 
>| Net:     lim   n[1- a(n+1)/a(n)]
>| 
>| Hopefully someone can clear this up for me.


Hi,

Well, your "Net" formula is not quite correct. Below you will find an
elaborate derivation of Raabe's test from which you may see that in your
"Net" formula the factor n outside the brackets should rather be n+1:

Let 1<s and

                                     1
                             v(n) = ----
                                      s
                                     n

Then

                    infinity        infinity
                     -----           -----
                      \               \       1
                       )     v(n) =    )     ----
                      /               /        s
                     -----           -----    n
                     n = 1           n = 1

converges since

    N
  -----
   \                 1      1        1      1      1      1
    )   v(n) = 1 + (---- + ----) + (---- + ---- + ---- + ----) + ...
   /                  s      s        s      s      s      s
  -----              2      3        4      5      6      7
  n = 1

or

               N
             -----
              \                    1           1
               )   v(n) < 1 + 2  (----) + 4  (----) + ...
              /                     s           s
             -----                 2           4
             n = 1

and

              N
            -----
             \                  1            1     2
              )   v(n) < 1 + -------- +  (--------)  + ...
             /                (s - 1)      (s - 1)
            -----            2            2
            n = 1

We set

                                    1
                             q = --------
                                  (s - 1)
                                 2

Since 1<s we have q<1. Thus the right-hand side of

                     infinity        infinity
                      -----           -----
                       \               \       n
                        )     v(n) <    )     q
                       /               /
                      -----           -----
                      n = 1           n = 0

converges and hence the left-hand side does so, too. Next we consider
the following identity:

                   /      v(n)  \     /    /n - 1\s\
                 n |1 - --------| = n |1 - |-----| |
                   \    v(n - 1)/     \    \  n  / /

If we rewrite this then we get in turn

                  /      v(n)  \                   s
                n |1 - --------| = n (1 - (1 - 1/n) )
                  \    v(n - 1)/

       /      v(n)  \     /                    s (s - 1)       \
     n |1 - --------| = n |1 -  (1 - s/n + 1/2 --------- - ...)|
       \    v(n - 1)/     |                        2           |
                          \                       n            /

              /      v(n)  \           s (s - 1)      1
            n |1 - --------| = s - 1/2 --------- + O(----)
              \    v(n - 1)/               n           2
                                                      n

                   /      v(n)  \
       lim       n |1 - --------| =
  n -> infinity    \    v(n - 1)/

                               s (s - 1)      1
             lim       s - 1/2 --------- + O(----)
        n -> infinity              n           2
                                              n

and finally

                                  /      v(n)  \
(1)                   lim       n |1 - --------| = s
                 n -> infinity    \    v(n - 1)/

Now we consider another series u(n). If u(n)>0 for all n and

                                  /      u(n)  \
(2)                   lim       n |1 - --------| = t ,
                 n -> infinity    \    u(n - 1)/

where 1<t, we can take

(3)                        s = 1/2  (t + 1) ,

which is greater than 1 but less than t, and then we can choose an m so
that for all n>m

                   /      v(n)  \     /      u(n)  \
                 n |1 - --------| < n |1 - --------|
                   \    v(n - 1)/     \    u(n - 1)/

since both sides converge (the left-hand side to (t+1)/2 and the
right-hand side to t). Rewriting the last inequality we have in turn

                   /    /n - 1\s\     /      u(n)  \
                 n |1 - |-----| | < n |1 - --------|
                   \    \  n  / /     \    u(n - 1)/

                           u(n)     /n - 1\s
                         -------- < |-----|
                         u(n - 1)   \  n  /

                                       /n - 1\s
                       u(n) < u(n - 1) |-----|
                                       \  n  /

and if we iteratively apply this recursion relation starting with u(m)
we see that for all n>m

                                           s
                          u(n) < u(m) (m/n)

Hence

                            infinity
                             -----
                              \
(4)                            )     u(n)
                              /
                             -----
                             n = 1

converges. Similarly, if the limit t in (2) exists and is less than 1,
the sum (4) diverges. Now

                             /      u(n)  \
                           n |1 - --------|
                             \    u(n - 1)/

becomes in turn:

                                /    u(n + 1)\
                        (n + 1) |1 - --------|
                                \      u(n)  /

                    /    u(n + 1)\         u(n + 1)
                  n |1 - --------| +  (1 - --------)
                    \      u(n)  /           u(n)

                u(n + 1)  /  u(n)      \         u(n + 1)
            n  (--------) |-------- - 1| +  (1 - --------)
                  u(n)    \u(n + 1)    /           u(n)

If

                                    u(n + 1)
(5)                       lim       -------- = 1
                     n -> infinity    u(n)

then we finally get Raabe's test:

                   /      u(n)  \                    /  u(n)      \
       lim       n |1 - --------| =      lim       n |-------- - 1|
  n -> infinity    \    u(n - 1)/   n -> infinity    \u(n + 1)    /

or

                                  /  u(n)      \
(6)                   lim       n |-------- - 1| = t
                 n -> infinity    \u(n + 1)    /


With kind regards,

Helmut