Assertion 2

There existsk₀ < ∞such that for allk > k₀ and 3⩽r<0.1kwe have

Proof

Note that this inequality is valid for all r ⩽ 11 and sufficiently large k. For r > 11 we have r² − (4/3)r + 2/3 < r² − (5/4)r and

If r > 11, then kr(r−1)<k8r, and for r<0.1k we have k8r>(5/4)k. Hence,

for all sufficiently large k.

Assertion 3

There existsk₀ < ∞such that for allk > k₀ and 3⩽r<0.1kthe following inequalities are valid:

Proof

In the first inequality the factors of the left hand side of (2lnr/r)(r2r2−2r+2)2 are bounded, so it is sufficient to show that the product of two other tends to zero uniformy over all r<0.1k as k tends to infinity. The relation

follows from the monotone descreasing of (r − 2)/(r² − r) with respect to r.

It is sufficient to show that for large k the value 2r3lnr(r−2r)k is small for all sufficiently large k:

Assertion 4

There existsk₀ < ∞ such that for all k > k₀ and 3⩽r<0.1kthe following inequality is valid:

Proof

It is sufficient to show that k4(r2−r−rk−1/8r(r−1))k tends to zero uniformly in r if k tends to infinity. The estimates

follows from the fact that the function r/(r2−r) is decreasing in r.

Assertion 5

There existsk₀ < ∞ such that for allk > k₀, 3⩽r<0.1k and arbitrary numbers δ_ij ∈ [0, r^−3/2k^−1/8] we have

Proof

Note that e−r−1/2<(1+r−1/2)−1=1−1r+1. Further,

So, the estimated value does not exceed

this expression for all sufficiently large k implies the desired upper bound 1−1/(2r).

Assertion 6

There existsk₀ < ∞ such that for allk > k₀, 3⩽r<0.1kthe following inequality is valid:

Proof

From the conditions on the parameters we deduce that

Further, r(rr+1)k⩽re−k/(r+1) tends to zero as k tends to infinity, hence it is smaller than 1/40 for large k. Therefore

where the last inequality also is satisfied for all sufficiently large k.

Assertion 7

There existsk₀ < ∞ such that for allk > k₀, 3⩽r<0.1kwe have

Proof

It is sufficient to show that each of the first three negative summands is 4 times smaller than the last summand. Let us compare each of them with the last summand. Begin with the first one:

for all sufficiently large k. Now consider the second summand:

It remains to consider the third summand:

for all sufficiently large k.

Assertion 8

There existsk₀ < ∞ such that for allk > k₀, 3⩽r<0.1kand arbitraryδii∈[0,(rr+1)k], i = 1, …, r, the following inequality is valid:

Proof

Note that lnδii<−kr+1 is sufficiently large in the absolute value, so for large k we may suppose that |δiilnδii+ln(rr−1)δii−δii|⩽2|δiilnδii|⩽2kln(r+1r)(rr+1)k⩽2kr+1(rr+1)k. Consequently, the left hand side of our inequality does not exceed

In view of condition on the relation between r and k the maximal summand in the second parenthesis is the first one. Hence, the left hand side of our inequality does not exceed

The assertion 8 is proved.