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### Mathematical Series

Date: 04/19/2005 at 14:43:52
From: James
Subject: Mathmatical series

I have a question relating to the transposition of a formula.  The
formula relates to reliability:

mtbf = 1/a + 1/2a + 1/3a + .... + 1/na

I can see that this is a mathematical series--I know mtbf is 23998 and
I also know a = 0.0008, but I am having difficulty finding n since it
changes at each stage and therefore I cannot put a direct value to it.

I could possibly single out n, but this is proving difficult because

Date: 04/20/2005 at 21:38:57
From: Doctor Vogler
Subject: Re: Mathmatical series

Hi James,

Thanks for writing to Dr. Math.  If you multiply both sides of your
equation

mtbf = 1/a + 1/2a + 1/3a + .... + 1/na

by the number a, then it becomes

mtbf*a = 1/1 + 1/2 + 1/3 + ... + 1/n.

This is much easier to work with, if you are trying to find n.  Now,
the simple way to find n is to multiply mtbf by a, and then subtract
off 1/1, and then subtract off 1/2, and then 1/3, and 1/4, and so on,
until you get something that is zero or less.  The number of fractions

That's the simple way.  But that's not the fast way.  In fact, if
mtbf*a is bigger than 19, then you'll need to subtract off over a
hundred million fractions before you get negative.  That's because

1/1 + 1/2 + 1/3 + ... + 1/n

is approximately the natural log of n,

ln n.

And so if you want

ln n > 19

then you need

n > e^19 > 100,000,000.

Of course, you should also notice that if you only have four or five
digits of precision on mtbf*a (and you actually only showed one digit
of precision on the number a), then you can't get the number n
exactly, since adding one more term

1/n

will only change the sum by less than 0.00000001.  And it is certain
that you will never get EXACTLY 23998*0.0008, so the best you can hope
for is a number n that gives you close to the right amount.

So what is the fast way to find n?  The fast way uses Euler's constant
gamma (that's a Greek letter that looks rather like a y with a loop on
the tail) and some approximations for the sum.

gamma = 0.5772156649015328606065120900824024310421593359399235988...

(You'll probably only need a few digits of this, like five or six, and
no more than ten.)  A good estimate that will certainly suffice for

ln (n+1) + gamma - 1/(2n) <
1/1 + 1/2 + 1/3 + ... + 1/n <
ln (n+1) + gamma - 1/(2n+2).

For example, if you know what you want the sum to equal

M = 1/1 + 1/2 + 1/3 + ... + 1/n,

then n will be approximately

n0 = e^(M - gamma),

and you can get better estimates by changing n0 to

e^(M - gamma + 1/(2*n0)).

For example, if you take

M = 23998*0.0008
= 19.1984

then you get

n0 = 122201800.06

and another iteration gives

122201800.56

So we take n = 122,201,800, and we find that the sum

1/1 + 1/2 + 1/3 + ... + 1/n

is between

19.1984000035373101405886978465851071841052388010642051815130

and

19.1984000035373101740709601813505258621221779810226505580255

But if we use one less term (n = 122,201,799), then the sum is between

19.1983999953541246228630992399870771093374538183247805496923

and

19.1983999953541246563453621227356287486853252543122747286441

For your information, I did these computations using the free program

http://pari.math.u-bordeaux.fr/

In Pari, the constant gamma is called Euler(), and so I asked it to
compute things like

n = 122201800
log(n+1) + Euler() - 1/(2*n)

By the way, if you need a better estimate for the sum (which is not
likely), then you can use

ln (n+1) + gamma - 1/(2n+2) - 1/12n(n+1) <
1/1 + 1/2 + 1/3 + ... + 1/n <
ln (n+1) + gamma - 1/(2n+2) - 1/12(n+1)(n+2).

back and show me what you have been able to do, and I will try to
offer further suggestions.

- Doctor Vogler, The Math Forum
http://mathforum.org/dr.math/

Date: 04/21/2005 at 16:31:13
From: James
Subject: Thank you (Mathmatical series)

Thanks very much for helping--it is very much appreciated.  It's nice
to know that there are people out there willing to help others.

James
Associated Topics:
High School Sequences, Series

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