two number-theoretical limits (& bonus sum)

[Leroy Quet]

Let d(k) = the number of positive divisors of k.

then, does:

limit{m -> oo}

(1/m) (sum{k=1 to m} d(k)) - ln(m) =

2*c -1,

where c = Euler's constant (.5772...)?

And related, does:

limit{m -> oo}

(1/m) sum{k=2 to m} (m(mod k))/k =

1 -c,

where 0 <= m(mod k) <= k-1 ?

I am doubting that these, at least the first, is true, but only
because the limit does not look familiar. (And such a limit *must* be
well-known.)

I derived ("derived") these limits from, in part,

sum{k=1 to m} sum{j|k, j <= sqrt(k)} a(j) =

sum{1 <= k <= sqrt(m)} a(k) (floor(m/k) +1 -k)

So, hopefully this sum.identity is correct anyhow.

(I set a(k) = 1, of course, to help get limits. I then found a limit
involving the number of positive divisors of k, where each divisor is
<= sqrt(k).)

(limit (1/m)(sum{k=1 to m} d'(k)) - ln(m)/2 = c -1/2,
where d'(k) = ceiling(d(k)/2) = # divisors <= sqrt(k).)

thanks,
Leroy Quet

[Martin Cohen]

Leroy Quet wrote:

The first part is correct. It is proved in good old Hardy and Wright,
theorem 320 in the form
d(1) + d(2) + ... + d(n) = n log n + 2 gamma - 1)n + O(sqrt(n)).

Don't know about part 2.

Martin Cohen

[Leroy Quet]

Martin Cohen <martin_...@raytheon.com> wrote in message news:<q4eob.119$b77...@dfw-service2.ext.raytheon.com>...

I presume you meant

d(1) + d(2) + ... + d(n) = n log n + (2 gamma - 1)n + O(sqrt(n))

(with an additional open-parenthesis), since you said my result was
correct.

:)

>
>Don't know about part 2.
>
>Martin Cohen

I used the identity

sum{k=1 to n} d(k) =

sum{k=1 to n} floor(n/k)

which is

sum{k=1 to n} (n - n(mod k))/k =

n*(1+1/2+...+1/n) - sum{k=1 to n} n(mod k))/k.

Since (1+1/2+...+1/n) - ln(n) -> Euler's constant, as n -> oo,
the second result in my original post follows.

(Note that the mod-sum is the same whether k starts at 1 or 2,
since the k=1 term is zero anyway.)

thanks,
Leroy Quet