Vtyjkyuk

I have a sequence of complex numbers $a_1,a_2,...$ such that $a_i neq -1$. I then have the infinite product $prodlimits_n=1^infty(1+a_n)$ which I know converges to a non zero complex number. I was wondering if this is enough to guarantee the convergence of $sumlimits_n=1^inftyfraca_n1+a_n$.

I couldn't come up with an obvious counterexample and my first idea for proving it was assuming $prodlimits_n=1^infty(1+a_n)=S$, so $sumlimits_n=1^inftyln(1+a_n)=ln S$ converges but have no idea how to deal with this since $1+a_i$ is complex.

If this diverges in general, I was wondering if there are any necessary and sufficient conditions for the sum to converge such as $sumlimits_n|a_n|<infty$ or that the $a_n$ are positive real numbers.

Any help would be awesome,and thanks for taking a look!

The infinite product $prod_n (1+a_n)$ converges (to a nonzero value) if and only if $sum_n log(1+a_n)$ converges (for a branch of $log$ that is analytic in a neighbourhood of $1$, with $log(1) = 0$). This does not in general imply $sum_n dfraca_n1+a_n$ converges. For example, take a positive sequence $b_k to 0 + $ and let $c_k = 1/(1+b_k) - 1$. Thus
$log(1+c_k) = -log(1+b_k)$. On the other hand,
$$ dfracc_k1+c_k + dfracb_k1+b_k = dfrac-b_k^21+ b_k ne 0$$
Consider a sequence $a_n$ that, for each $k$, repeats $b_k, c_k$ enough times for the partial sum of $a_n/(1+a_n)$ to change by $1$, before progressing to the next $k$. Then $sum log(1+a_n)$ converges to $0$ but
$sum a_n/(1+a_n)$ diverges.

It's OK for absolute convergence. But not in general.

Take $log(1+a_n) = (-1)^n/sqrtn$ for $n ge 1$, so
$sum log(1+a_n) = sum (-1)^n/sqrtn$ converges (conditionally). But
$$
fraca_n1+a_n = frac(-1)^nsqrtn - frac12n +Oleft(frac1n^3/2right)
$$
and thus $sum a_n/(1+a_n)$ diverges.