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Consider the divisors of $n$, $$d_1 = 1, d_2, d_3, ..., d_r=n$$ in ascending order and $r \equiv r(n)$ is the number of divisors of $n$.

Is there any expression $f(n) < r(n)$ such that, $$\sum_{k=1}^{r-1} \frac{d_k}{d_{k+1}} < f(n)$$

Or equivalently, in a generalized way, is there any expression $\phi_a(n) < \sigma_a(n)$ such that, $$\sum_{k=1}^{r-1} \frac{d_k^{a+1}}{d_{k+1}} < \phi_a(n)$$ for positive integers $a$ ?

PS. Looking for a study on tight bounds over these sums of ratios. Any references will be greatly appreciated.

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    I added the reference-request tag to your question due to your post-script.2011-12-30

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This is not a complete answer but I would like to share some thoughts. First I noticed that $S(p^n)=n/p$, where $S$ is your first sum. This means that any good bound will probably depend on knowledge of the distribution of prime factors in $n$. Denote from now on the $N$-th prime with $p_n$.

Now let's consider $n=pq$ where $p

So the $S(p_n p_{n+1})$ numbers converge on 1 but are not strictly smaller then 1. You can derive an explicit bound out of this if you want to.

Now let's consider $n=pqр$ where $pr$ and $S(pqr)=4/p + 2p/q + pq/r$ when $pq

Numerical experiments showed that $S(\prod_{i=1}^n{q_i}$), where $q_i$ are primes is convergent to the Eulerian number $A(n,1)$. Please note that I can't prove that.

There seem to be similar limits for all divisor multiplicities. I will edit this post if I manage to glean the logic behind it.

EDIT:

Let $M(r)$ be the set of all natural numbers which have multiplicity of their prime divisors $r=(r_1,r_2,..r_n)$. Let $s_i$ be the i-th symmetric polynomial of n variables.

Numerical simulations seem to show that $$\lim_{x\to\infty, x\in M}S(x)=\sum_{i=2}^n{s_i(r)}$$ Note that the summation starts from 2 - i.e. the second symmetric polynomial.

This may be possible to prove by induction if the initial cases are established. Consider a number $X$ with divisors $1=d_1

This is of course no proof at all, just some thoughts on the subject.