Divergence of a Recurrence Relation with the Divisor Function Define the recurrence relation {$a_{n}$} as so: 
$$a_{n+1}=\tau(\sum _{ i=1 }^{ n }{ a_{ i } })$$
Where $\tau (n) =\sigma_0 (n)$, and $\sigma_k (n)$ is the divisor function. 
For example, if $a_1=6$, the sequence {$a_{n}$} is $6,4,4,4,6,8,6,4,…\dots$
However, is {$a_{n}$} is convergent or divergent? 
I thought that it would be divergent. However, while $\sum _{ i=1 }^{ n }{ a_{ i } }$ is always increasing, $\tau (n)$ is not a increasing function, so the problem proved difficult to solve.
I would appreciate any help. 
 A: The sequence does not converge. Since it is an integer sequence, if it were convergent, it would be eventually constant, say $a_n = c$ for $n \geqslant N > 1$. Since the values of $\tau$ are strictly positive, we'd have $c > 0$. Let $A = \sum_{n = 1}^{N-1} a_n$. Under the assumption that $a_n = c$ for $N \leqslant n < N + A$ we find
$$a_{N+A} = \tau(A + A\cdot c) = \tau\bigl((c+1)A\bigr) > \tau(A).$$
Thus $(a_n)$ cannot be eventually constant, hence the sequence is divergent.
An interesting question is whether $(a_n)$ is unbounded.
A: I'll try to prove that  your sequence is unbounded.
write $b_k = \sum_{n=1}^k a_n$ with $a_{n+1} = \tau(b_n)$ and suppose that the $a_n$ are bounded by $C$.
the mean distance between two consecutive primes tends to be greater and greater, i.e. $\sum_{p \le x} 1 = o(x)$,
in the same way, with $E_C = \{ n \in \mathbb{N}^* \, |\, \tau(n) \le C \}$ the mean distance between two consecutive elements becomes greater and greater : $$\sum_{b \in E_C, \ \ b \le x} 1 = o(x)$$

to prove this, just consider the series $\displaystyle(\sum_{p\in \mathcal{P}} p^{-\sigma})^C = \sum_{k =1}^\infty u_C(k) k^{-\sigma}$ for $\sigma \in ]1;\infty[$ and note that all the coefficients $u_C(k) \ge 0$.
obviously that sequence $u_C(k)$ contains the set $E_C$ in the sense that if $b \in E_C$ then $u_C(b) \ge 1$, so that if $\sum_{k \le x} u_C(k) = o(x)$ then $\displaystyle\sum_{b \in E_C, \ \ b \le x} 1 \le \sum_{k=1}^x u_C(k) = o(x)$.
and because when $\sigma \to 1^+$ : $\sum_{p\in \mathcal{P}} p^{-\sigma} \sim \ln \zeta(\sigma) \sim |\ln(\sigma-1)|$ we get that $(\sum_p p^{-\sigma})^C \sim |\ln(\sigma-1)|^C = o(\frac{1}{\sigma-1})$ which implies $\sum_{k \le x} u_C(k) = o(x)$
(that's a theorem for Dirichlet series with non-negative coefficients $(a_n)$ that the pole on the real axis with the largest real part determinates the growth rate of $\sum_{n \le x} a_n$).

now  suppose that $a_{n+1} = \tau(b_n)\le C$ we get $b_n \in E_C$ hence $\displaystyle\lim \sup_n a_n = \displaystyle\lim \sup_n b_n - b_{n-1} \to \infty$, i.e. your $a_n$ are unbounded
