Conjecture about $ f(x,v) = 1 + \sum_{n=1}^\infty {x^n \over t(v n) } > 0 $ for all real $x$

Question

Let $t(1) = 1$ and $t(x)$ be strictly increasing for $x>0$. Also $0 \leq t(x)$ for $0 \leq x$ and $t(z)$ is meromorphic on the entire complex plane.

Conjecture :

Let $f(z,v)$ be analytic in $z$ for all complex $z$ (entire function) and all real $0 \leq v$.

If $ f(x,v) = 1 + \sum_{n=1}^\infty {x^n \over t(v n) } > 0 $ for all real $x$ and $0<v<1/3$ and $2/3<v<1$ then

$ f(x,v) = 1 + \sum_{n=1}^\infty {x^n \over t(v n) } > 0 $ for all real $x$ and $1/3<v<2/3$.

Some motivation/background/inspiration

Solve for $m(t)$ in the integral transform $\int_0^1(1-t^n) m(t) dt=\frac{(n+1)^n}{n^{n-1}} $ for $n>0$.

$0 \leq C + \sum_{n=1}^{+\infty} \frac{x^n}{(n+1)^n} $ ,where $C = 1.036055393...$ is the best possible constant.

And most of all

Is $ f(x,v) = \sum_{n=0}^\infty {x^n \over \Gamma(v n +1) } > 0 $ for all real $x$ and $0<v<1$?

where "t" is the factorial.

It felt natural to consider changing $t(n)$ into $t(vn)$. Notice the analogue changing $x^n$ into $x^{vn}$ always works.

So it felt a bit like a " substitution for integrals " but for a sum here.

Answer 1

If $t(vn)=vn$, for $v\ne 0$ we have that $$ f(x,v) = 1+ \frac{1}{v} \sum_{n \ge 1}\frac{x^n}{n}, $$

where the series converges for $|x|<1$. So, in general, $f(x,v)$ may not even be defined for all $x$. So I'd say the conjecture does not hold with the present conditions.