## 4 limits

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• Who can help me calculating up to 10 exact decimal places sum (1/n^(1-1/n) - 1/n) sum (1/p^(1-1/p) - 1/p) sum (ln(n) / n^2) sum (ln(p) / p^2) n=positive
Message 1 of 9 , Jul 4, 2007
Who can help me calculating up to 10 exact decimal places

sum (1/n^(1-1/n) - 1/n)

sum (1/p^(1-1/p) - 1/p)

sum (ln(n) / n^2)

sum (ln(p) / p^2)

n=positive integer, p=prime, each from 1(2) to infinity?

WDS

[Non-text portions of this message have been removed]
• Thanks for evaluations received. What I mean is: Can sum(ln(n)/n^2 ~ 0.93 reach 1? Can sum(ln(p)/p^2 ~ 0.49 reach 0.5? WDS ... [Non-text portions of this
Message 2 of 9 , Jul 7, 2007
Thanks for evaluations received. What I mean is:

Can sum(ln(n)/n^2 ~ 0.93 reach 1?

Can sum(ln(p)/p^2 ~ 0.49 reach 0.5?

WDS

--- In primenumbers@yahoogroups.com, "Werner D. Sand" <Theo.3.1415@...>
wrote:
>
>
> Who can help me calculating up to 10 exact decimal places
>
>
>
> sum (1/n^(1-1/n) - 1/n)
>
> sum (1/p^(1-1/p) - 1/p)
>
> sum (ln(n) / n^2)
>
> sum (ln(p) / p^2)
>
>
>
> n=positive integer, p=prime, each from 1(2) to infinity?
>
>
>
> WDS

[Non-text portions of this message have been removed]
• Hello, ... Consider the inequality sum(ln(k)/k^2, k=A+1..infinity)
Message 3 of 9 , Jul 9, 2007
Hello,

> Can sum(ln(n)/n^2 ~ 0.93 reach 1?

Consider the inequality

sum(ln(k)/k^2, k=A+1..infinity) < integral(ln(k)/k^2, k=A..infinity)

The right-hand side can be evaluated as (ln(A)+1)/A. Taking A=2 yields
sum(ln(k)/k^2, k=4..infinity) < (ln(3)+1)/3 < 0.700
Clearly,
sum(ln(k)/k^2, k=1..3) = ln(2)/4 + ln(3)/9 < 0.296

Summing these two inequalities shows that your sum is bounded above by
0.996. This estimate is quite rough, though; the actual value of the
infinite sum seems to be around 0.9375...

> Can sum(ln(p)/p^2 ~ 0.49 reach 0.5?

Unfortunately, I don't know a nice trick for this one :-)

Peter

--
[Name] Peter Kosinar [Quote] 2B | ~2B = exp(i*PI) [ICQ] 134813278
• Using a similar trick, one can verify the sum over the primes is strictly less than 0.5. One overestimates the tail end of the sum: sum(log(t)/t^2,t=p to 00,
Message 4 of 9 , Jul 12, 2007
Using a similar trick, one can verify the sum over the primes is
strictly less than 0.5. One overestimates the tail end of the sum:
sum(log(t)/t^2,t=p to 00, t prime) > sum(log(t)/t^2,t=p to 00, all t,
prime or not) > int(log(t)/t^2,t=p-1 to 00, any t). If you use
p=1009 then the initial segment sum is
about .4921002678856370264032818 while the error term (integral) is
upper bound of) .4999531681322950759326119. So the sum over all
primes is strictly less than 0.5.

--- In primenumbers@yahoogroups.com, Peter Kosinar <goober@...> wrote:
>
> Hello,
>
> > Can sum(ln(n)/n^2 ~ 0.93 reach 1?
>
> Consider the inequality
>
> sum(ln(k)/k^2, k=A+1..infinity) < integral(ln(k)/k^2,
k=A..infinity)
>
> The right-hand side can be evaluated as (ln(A)+1)/A. Taking A=2
yields
> sum(ln(k)/k^2, k=4..infinity) < (ln(3)+1)/3 < 0.700
> Clearly,
> sum(ln(k)/k^2, k=1..3) = ln(2)/4 + ln(3)/9 < 0.296
>
> Summing these two inequalities shows that your sum is bounded above
by
> 0.996. This estimate is quite rough, though; the actual value of
the
> infinite sum seems to be around 0.9375...
>
> > Can sum(ln(p)/p^2 ~ 0.49 reach 0.5?
>
> Unfortunately, I don't know a nice trick for this one :-)
>
> Peter
>
> --
> [Name] Peter Kosinar [Quote] 2B | ~2B = exp(i*PI) [ICQ]
134813278
>
• Whoops! Of course those inequalities are the other way around!! One overestimates the tail end of the sum: sum(log(t)/t^2,t=p to 00, t prime)
Message 5 of 9 , Jul 13, 2007
Whoops!

Of course those inequalities are the other way around!!

One overestimates the tail end of the sum:
sum(log(t)/t^2,t=p to 00, t prime) < sum(log(t)/t^2,t=p to 00, all t,
prime or not) < int(log(t)/t^2,t=p-1 to 00, any t).... etc.

>
> Using a similar trick, one can verify the sum over the primes is
> strictly less than 0.5. One overestimates the tail end of the sum:
> sum(log(t)/t^2,t=p to 00, t prime) > sum(log(t)/t^2,t=p to 00, all
t,
> prime or not) > int(log(t)/t^2,t=p-1 to 00, any t). If you use
> p=1009 then the initial segment sum is
> about .4921002678856370264032818 while the error term (integral) is
> upper bound of) .4999531681322950759326119. So the sum over all
> primes is strictly less than 0.5.
>
>
> --- In primenumbers@yahoogroups.com, Peter Kosinar <goober@> wrote:
> >
> > Hello,
> >
> > > Can sum(ln(n)/n^2 ~ 0.93 reach 1?
> >
> > Consider the inequality
> >
> > sum(ln(k)/k^2, k=A+1..infinity) < integral(ln(k)/k^2,
> k=A..infinity)
> >
> > The right-hand side can be evaluated as (ln(A)+1)/A. Taking A=2
> yields
> > sum(ln(k)/k^2, k=4..infinity) < (ln(3)+1)/3 < 0.700
> > Clearly,
> > sum(ln(k)/k^2, k=1..3) = ln(2)/4 + ln(3)/9 < 0.296
> >
> > Summing these two inequalities shows that your sum is bounded
above
> by
> > 0.996. This estimate is quite rough, though; the actual value of
> the
> > infinite sum seems to be around 0.9375...
> >
> > > Can sum(ln(p)/p^2 ~ 0.49 reach 0.5?
> >
> > Unfortunately, I don't know a nice trick for this one :-)
> >
> > Peter
> >
> > --
> > [Name] Peter Kosinar [Quote] 2B | ~2B = exp(i*PI) [ICQ]
> 134813278
> >
>
• Fine trick, but what makes you sure the integral is not e.g. = 0.008? Didn t you only shift the problem upon the integral? Werner ... [Non-text portions of
Message 6 of 9 , Jul 14, 2007
Fine trick, but what makes you sure the integral is not e.g. = 0.008?
Didn't you only shift the problem upon the integral?

Werner

>
> Whoops!
>
> Of course those inequalities are the other way around!!
>
> One overestimates the tail end of the sum:
> sum(log(t)/t^2,t=p to 00, t prime) < sum(log(t)/t^2,t=p to 00, all t,
> prime or not) < int(log(t)/t^2,t=p-1 to 00, any t).... etc.
>
>
> >
> > Using a similar trick, one can verify the sum over the primes is
> > strictly less than 0.5. One overestimates the tail end of the sum:
> > sum(log(t)/t^2,t=p to 00, t prime) > sum(log(t)/t^2,t=p to 00, all
> t,
> > prime or not) > int(log(t)/t^2,t=p-1 to 00, any t). If you use
> > p=1009 then the initial segment sum is
> > about .4921002678856370264032818 while the error term (integral) is
> > upper bound of) .4999531681322950759326119. So the sum over all
> > primes is strictly less than 0.5.
> >
> >
> > --- In primenumbers@yahoogroups.com, Peter Kosinar <goober@> wrote:
> > >
> > > Hello,
> > >
> > > > Can sum(ln(n)/n^2 ~ 0.93 reach 1?
> > >
> > > Consider the inequality
> > >
> > > sum(ln(k)/k^2, k=A+1..infinity) < integral(ln(k)/k^2,
> > k=A..infinity)
> > >
> > > The right-hand side can be evaluated as (ln(A)+1)/A. Taking A=2
> > yields
> > > sum(ln(k)/k^2, k=4..infinity) < (ln(3)+1)/3 < 0.700
> > > Clearly,
> > > sum(ln(k)/k^2, k=1..3) = ln(2)/4 + ln(3)/9 < 0.296
> > >
> > > Summing these two inequalities shows that your sum is bounded
> above
> > by
> > > 0.996. This estimate is quite rough, though; the actual value of
> > the
> > > infinite sum seems to be around 0.9375...
> > >
> > > > Can sum(ln(p)/p^2 ~ 0.49 reach 0.5?
> > >
> > > Unfortunately, I don't know a nice trick for this one :-)
> > >
> > > Peter
> > >
> > > --
> > > [Name] Peter Kosinar [Quote] 2B | ~2B = exp(i*PI) [ICQ]
> > 134813278
> > >
> >
>

[Non-text portions of this message have been removed]
• ... The integral can be evaluated explicitly, exactly as I did in my original post. I just didn t find this method nice trick -y enugh, as it required adding
Message 7 of 9 , Jul 14, 2007
> Fine trick, but what makes you sure the integral is not e.g. = 0.008?
> Didn't you only shift the problem upon the integral?

The integral can be evaluated explicitly, exactly as I did in my original
post. I just didn't find this method "nice trick"-y enugh, as it required
adding 168 terms first (as opposed to the two terms in the other sum).
It's easy to check that the integral int(log(t)/t^2, t=A..infinity) is
equal to (ln(A)+1)/A.

Adam first added all the terms corresponding to primes smaller than 1000,
which resulted in roughly 0.4921003 and then bounded the remaining terms
by the value of the integral corresponding to p=1009 (which means
A = p-1 = 1008). It's easy to see that (ln(1008)+1)/1008 < 0.007853.
These two values add up to 0.4999533, which is strictly less than 0.5.

Peter
• ... 0.008? ... original ... required ... sum). ... is ... 1000, ... terms ... 0.5. ... You are quite right. I integrated only numerically. Everything clear.
Message 8 of 9 , Jul 16, 2007
--- In primenumbers@yahoogroups.com, Peter Kosinar <goober@...> wrote:
>
> > Fine trick, but what makes you sure the integral is not e.g. =
0.008?
> > Didn't you only shift the problem upon the integral?
>
> The integral can be evaluated explicitly, exactly as I did in my
original
> post. I just didn't find this method "nice trick"-y enugh, as it
required
> adding 168 terms first (as opposed to the two terms in the other
sum).
> It's easy to check that the integral int(log(t)/t^2, t=A..infinity)
is
> equal to (ln(A)+1)/A.
>
> Adam first added all the terms corresponding to primes smaller than
1000,
> which resulted in roughly 0.4921003 and then bounded the remaining
terms
> by the value of the integral corresponding to p=1009 (which means
> A = p-1 = 1008). It's easy to see that (ln(1008)+1)/1008 < 0.007853.
> These two values add up to 0.4999533, which is strictly less than
0.5.
>
> Peter

You are quite right. I integrated only numerically. Everything clear.
Thanks to you both.

Werner
>
• Hi, I m not sure if my email got through to you so I ll post here. I put a message on usenet a few years back
Message 9 of 9 , Jul 28, 2007
Hi, I'm not sure if my email got through to you so I'll post here.
I put a message on usenet a few years back

with possible expansions for some prime sums, all unproven!
You should also check Henri Cohen's paper (a dvi file)
http://www.math.u-bordeaux.fr/~cohen/hardylw.dvi
which has some evaluations, especially the last one!

Andrew

--- In primenumbers@yahoogroups.com, "Werner D. Sand"
<Theo.3.1415@...> wrote:
>
>
> Who can help me calculating up to 10 exact decimal places
>
>
>
> sum (1/n^(1-1/n) - 1/n)
>
> sum (1/p^(1-1/p) - 1/p)
>
> sum (ln(n) / n^2)
>
> sum (ln(p) / p^2)
>
>
>
> n=positive integer, p=prime, each from 1(2) to infinity?
>
>
>
> WDS
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
> [Non-text portions of this message have been removed]
>
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