= Solution
First, for $0<x<1$, the <derivative> of $h(x)=\log(1-x)+x$ is $-x/(1-x)<0$, while $h(0)=0$. Hence $\log(1-x)<-x$.
Choose $N>c$ large enough that the given ratio bound holds for every $n\ge N$. Positivity permits taking <logarithms>, and the preceding inequality gives
$$
\log a_{k+1}-\log a_k\le\log(1-c/k)<-c/k.
$$
Summing, and using the <integral test for convergence> from part (b), gives for $n>N$
$$
\log a_n\le\log a_N-c\sum_{k=N}^{n-1}\frac1k
\le\log a_N-c\log(n/N).
$$
Thus $a_n\le a_NN^c n^{-c}$. Since $c>1$, the <P-series> $\sum n^{-c}$ converges, so the <comparison test for series> proves
$$
\boxed{\sum_{n=1}^\infty a_n<\infty.}
$$
The finitely many initial terms do not affect convergence. This <power-law comparison from a refined ratio bound> supplies information when the ordinary <ratio test> can be inconclusive.
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