Sum of sequence of 1/n
Mike Housky
[emailed and posted to alt.algebra.help]
Dan Larsen wrote:
>
> According to the nth limit test, as n-> oo, and the limit approaches 0,
> then sum of the sequence converges, and can be found.
Not quite. This is a one-way test. If the terms of a series (the members
of the sequence) don't converge to zero as a limit, the series does not
converge. The infinite sum is not defined. However, the test doesn't
work the other way. If the limit of the sequence exists and is zero,
this doesn't guarantee that the sum of the sequence is defined.
> If you were to take the sequence (2n^2+10)/(n^2-10), the series converges
> to 2, and the sum diverges. This is evident that by adding terms where n
> is very large, you basically keep adding 2. However, why does the sum of
> the series 1/n diverge? the limit converges to 0, and doing a basic test
> says that as n becomes large, you are basically adding zero, which means
> you don't add anything.
The "series 1/n" is the classic counterexample. It's called the Harmonic
Series because the sequence (1,1/2,1/3,1/4,...) is intimately related
to musical harmony. For example, these are the wavelengths of an ideal
plucked string of length=1.
The only proof that I know offhand that this series diverges is
based upon comparing the sequence of partial sums:
H[n] = Sum[i=1 to n](1/n)
to the natuaral logarithm of n, which is defined to be:
ln(x) = Integral[t=1 to x](dt/t)
for positive real x. For x>1, this is the area under the curve y=1/x
between 1 and x. What can be shown is that H[n] > ln(n) for all n>=1,
and this is done by converting the H[n] to the sum of areas of rectangles
that can be stacked side-by-side to form a structure whose value is
greater than or equal to ln(n) for every n>1.
A piece of graph paper will help. Draw curve A as y=A(x)=1/x for x>=1.
Draw curve B as 1/floor(x), where floor(x) is x if x is an integer, or
the next smaller integer if x is not an integer. Look at the area
under each rectangle. If you've drawn these correctly--translation: as
I've imagined--then you will see that curve B is always greater than or
equal to curve A when x is greater than or equal to one. The obvious
conclusion is that if ln(n) diverges as n->+inf then so does H[n].
A harder problem is proving that ln(x) as defined above is indeed a
logarithm function. I'll leave that for you to uncover during your
future studies. The thing to prove is that ln(xy) = ln(x) + ln(y)
for all positive x,y. That, and the obvious fact that ln(x) is
positive for x>1, makes it easy to prove that ln(x) is unbounded.
Cheers,
Mike.
glenn
(Mailed and posted)
Dan Larsen wrote:
>
> According to the nth limit test, as n-> oo, and the limit approaches 0,
> then sum of the sequence converges, and can be found.
> If you were to take the sequence (2n^2+10)/(n^2-10), the series converges
> to 2, and the sum diverges. This is evident that by adding terms where n
> is very large, you basically keep adding 2. However, why does the sum of
> the series 1/n diverge? the limit converges to 0, and doing a basic test
> says that as n becomes large, you are basically adding zero, which means
> you don't add anything.
>
> -Dan
This diverges because as n becomes large, you are still adding
something. Look at it this way. After 1/1, 1/2 the sum is 1 1/2.
Notice that the next two terms are => than 1/4. thus two of them adds
at least 1/2 to the sum. The next four terms are 1/5, 1/6, 1/7, and
1/8. They are => to 1/8, thus adding four of them increases the sum by
at least 1/2. Taking twice as many terms each time increases the sum by
at least 1/2. Thus there are an infinite collection of numbers, all of
which are at least 1/2, thus the sum diverges. ~Glenn~
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Mike Hill
The nth term test is the easiest test to use. However, it doesn't catch
sequences like 1/n that actually diverge. All that the test says is that if the
sequence converges, then lim(n->oo)a_n=0. Some sequences whose lim is 0 do not
converge. For 1/n, use the integral test:
/oo
|1/n dn=lim(n->oo) ln(n)-ln(1)=oo, the integral diverges, so the sequence
/1 diverges.
In article <01bc56ae$78dbcbe0$1fbc...@century.centuryinter.net>,
danl...@centuryinter.net says...
DownEasta
There is an old "bar trick" concerning quick sum of the numbers (integer)
from 1
to N. It is clamed that a child thought to be "dull" answered immediately
when his 3rd grade teacher asked "what is the sum of the numbers from 1 to
10?"
The "dull" kid quickly raised his hand and said "55." The teacher then
tried some other sums from 1 to N and the kid answered nearly as quickly.
The "legend" may be true, but the method is simple.
If N is odd remember N. Divide N-1 (next lower even integer) by 2
if N is even divide N by 2 eg 10/2=5
Multiply the result (in this case 5) by the sum of the first number (1)
and N (10 in
this example) 5x11=55
If N was odd, then just add N to the total eg: N=11, N is odd so {N-1}=10
and 10/2=5, 5X10=55, 55+11 {N} =66
Our hypothetical "dull" kid had said 1+10 is eleven and 2+9 is eleven and
3+8=11.... so if I take half of the number of numbers and multiply that by
the sum of the first plus the last, I'll have the answer.
Lets try - the sum of the numbers from 1 to 27
27 is odd so we divide {N-1}=26 by 2 giving 13 (27-1)/2=13
13x26=(10x26)+(3x26)=260+78=300+28=328 just the way I do it...
and 328 + 27 {the odd remainder} = 355
Start with low numbers and do it in your gray-meat computer.
The number is 8
The sum of the numbers from 1 to 8 is
9{ one plus last} times 4 {half of 8)
equals 36
Now try 1 to 9
remember N=9, make it even N1=8 do as above and add the 9
36+9=45
Well, I had fun sending this reply. Hope you did too!
Andy Eliason <ahel...@aol.com>
Pilpel Haran
In alt.algebra.help DownEasta <down...@aol.com> wrote:
<snip>
> The "dull" kid quickly raised his hand and said "55." The teacher then
<lots more snipping>
Well, actually (IIRC), the 'dull' kid was Gauss...
> Andy Eliason <ahel...@aol.com>
Haran
Vincent Johns
(posted & emailed)
DownEasta <down...@aol.com> wrote:
>
> There is an old "bar trick" concerning quick sum of the numbers (integer)
> from 1
> to N. It is clamed that a child thought to be "dull" answered immediately
> when his 3rd grade teacher asked "what is the sum of the numbers from 1 to
> 10?"
>
> The "dull" kid quickly raised his hand and said "55." The teacher then
> tried some other sums from 1 to N and the kid answered nearly as quickly.
> The "legend" may be true, but the method is simple.
> [...]
This sounds like the story (recounted by Eric Temple Bell)
about K. F. Gauss at the age of about 8 years, except that
probably nobody considered Gauss to be "dull", just not
yet at that age a great mathematician.
As I recall the story, Herr Büttner, the teacher, had given
the boys in the class about an hour to add up a set of 100 numbers
such as 5192 + 5229 + 5266 + ... , where each one was 37 larger
than the previous one. I don't know the starting number nor the
increment, but they formed an arithmetic progression, the kids
were probably supposed to derive each term before adding it, and
the teacher had a secret formula for determining the answer.
My guess is that Gauss figured out that the teacher had access
to something he wasn't sharing and independently derived a
slick way to find the sum, by rearranging the order of summing.
Maybe it wasn't exactly divine inspiration, but it still took
a pretty impressive mind to come up with that technique at that
age.
Gauss just wrote the answer on his slate (no calculations), and
he and Büttner sort of glared at each other for an hour while
the other boys slaved away. Gauss later said that his answer
was the only correct one turned in that day.
The story has a happy ending -- the teacher, recognizing that
there wasn't much more that he could teach this unusual student,
arranged for a tutor to take charge of Gauss's education, and the
tutor and Gauss became lifelong friends and collaborators.
--
-- Vincent Johns
Please feel free to quote anything I say here.
BrianScott
har...@post.tau.ac.il (Pilpel Haran) wrote:
>In alt.algebra.help DownEasta <down...@aol.com> wrote:
><snip>
>> The "dull" kid quickly raised his hand and said "55." The teacher
then
><lots more snipping>
>Well, actually (IIRC), the 'dull' kid was Gauss...
And the series was quite a bit uglier. I seem to remember a rather
large number of terms and a constant difference greater than 1, not
to mention the fact that the terms were 6-digit numbers or thereabouts.
Brian M. Scott
Do Not Use: brian...@aol.com
Always Use: sc...@math.csuohio.edu