It's been noted already in the comments that the problem is still
too easy as stated, because one can easily find functions
$f(z) = \sum_{n=0}^\infty c_n z^n$ such as $f(z) = \sin(1 / (r-z))$,
with infinitely many zeros inside the circle of convergence $|z|<r$,
and as long as $r>1$ the coefficients $c_n$ decay exponentially,
and thus satisfy constraints such as $\sum_n n|c_n| < \infty$
or $\left|c_n-c_{n-1}\right| \ll n^{-3}$ with plenty of room to spare.

[It's apparently not been noted that the "nonpolynomial" condition
is automatic unless $f$ is identically zero, since a nonzero polynomial
has finitely many roots anywhere in ${\bf C}$.]

However, the problem becomes somewhat more interesting if we
require (as Johan Andersson suggested, and as the proposer
may have intended) that $f$ have infinitely many roots
in the *unit* disc $|z|<1$, which is the largest disc on which
the power series is guaranteed to converge by the constraints on $c_n$.

Still, even with that interpretation there are power series
whose coefficients that satisfy either condition, or even
$|c_n - c_{n-1}| \ll n^{-k}$ for any constant $k$,
and nevertheless have infinitely many *real* zeros in $\left|z\right|<1$.
I'll deal with $\left|c_n - c_{n-1}\right| \ll n^{-3}$,
though the same technique applies for all $k$.
Since $c_n - c_{n-1}$ is the $z^n$ coefficient of $(1-z) \phantom. f(z)$,
it is enough to construct $g(z) = \sum_{n=0}^\infty a_n z^n$
with $|a_n| \ll n^{-3}$ and infinitely many real zeros,
and then set $f(z) = g(z) / (1-z)$.

Define $g(z) = S(z) - P(z)$, where
$$
S(z) = z
- \frac{z^2}{2^3}
+ \frac{z^4}{4^3}
- \frac{z^8}{8^3}
+ \frac{z^{16}}{16^3} - + \cdots
= \sum_{m=0}^\infty (-1)^m \frac{z^{2^m}}{2^{3m}},
$$
and $P(z)$ is the cubic polynomial
$$
P(z) = \frac89 - \frac45(1-z) - \frac1{15} (1-z)^2 - \frac1{60}(1-z)^3.
$$
Clearly $S$ satisfies the functional equation $S(z) = z - \frac18 S(z^2)$.
The cubic $P$ was chosen to satisfy the same equation within $O((1-z)^4)$:
we have
$$
P(z) = z - \frac18 P(z^2) + (1-z)^4 \frac{(1+z)(3+z)}{480}.
$$
Hence $g(z) = -\frac18 g(z^2) + O((1-z)^4)$. It follows that
as $z \rightarrow 1$ the ratio $f(z) / (\log(1/z))^3$
approaches some function $\phi(z)$
that satisfies $\phi(z^2) = -\phi(z)$ exactly.
Unless this function is identically zero, it must have
at least one zero in each interval $(z,z^{1/2}]$, and thus
an infinite sequence of zeros approaching $1$.
Now there are several ways to show that $\phi$ is not
identically zero, but possibly the simplest is numerical
computation, which shows $\phi$ oscillating with amplitude
about $2.2 \cdot 10^{-5}$. The first zero of $f(z)$ appears near $z=0.998131$,
then $0.998891$, $0.999493$, $0.999736$, $0.999871$, $0.999935$, $0.999967$,
etc. with each zero about twice as close to $1$ as the previous one.

Here's some **gp** code that gives a rough plot showing $f$ first
crossing zero after not quite reaching it around $0.9945$
(so it's 99.44% certain to be negative?...), and then locates
the first seven zeros exhibited above.

```
f(z) = suminf(n=0,(-1)^n*z^2^n/8^n)
P(z) = 8/9 - (1-z)*4/5 - (1-z)^2/15 - (1-z)^3/60
R(z) = (f(z)-P(z)) / log(1/z)^3
plot(x=.99,.9993,R(x))
for(i=11,17,print(solve(x=1-3/(2^(i-1)),1-3/2^i,R(x))))
```