The situation is governed by the following instance of a very general comparison principle (so general that I wouldn't even try to formulate it in full).

Let $f:[0,+\infty)\to[0,+\infty)$ be a decreasing function. Let $a_0=0$, $a_k=a_{k-1}+f(a_{k-1})$. Let $A(0)=0$, $A'(x)=f(A(x))$. Then for all $k$, we have $|a_k-A(k)|\le f(0)$.

Proof:

Let $k$ be the first index for which $a_k>A(k)+f(0)$ (if it exists). Then $a_{k-1}=a_k-f(a_{k-1})\ge a_k-f(0)>A(k)$, so $a_k-a_{k-1}\le f(A(k))$ while $A(k)-A(k-1)=\int_{k-1}^k f(A(t))\,dt\ge f(A(k))$, so we must already have $a_{k-1}>A(k-1)+f(0)$, contradicting the choice of $k$.

Similarly, let $k$ be the first index for which $a_k<A(k)-f(0)$ (if it exists). Then $A(k-1)=A(k)-\int_{k-1}^k f(A(t))\,dt\ge A(k)-f(0)>a_k$, so $A(k)-A(k-1)=\int_{k-1}^k f(A(t))\,dt\le f(a_k)$ while $a_k-a_{k-1}=f(a_{k-1})\ge f(a_k)$, so we must already have $a_{k-1}<A(k-1)-f(0)$, contradicting the choice of $k$.

Now just solve the related differential equation and use the fact that $\log(1-1/n)\asymp 1/n$ for large $n$.

The situation is governed by the following instance of a very general comparison principle (so general that I wouldn't even try to formulate it in full).

Let $f:[0,+\infty)\to[0,+\infty)$ be a decreasing function. Let $a_0=0$, $a_k=a_{k-1}+f(a_{k-1})$. Let $A(0)=0$, $A'(x)=f(A(x))$. Then for all $k$, we have $|a_k-A(k)|\le f(0)$.

Proof:

Let $k$ be the first index for which $a_k>A(k)+f(0)$ (if it exists). Then $a_{k-1}=a_k-f(a_{k-1})\ge a_k-f(0)>A(k)$, so $a_k-a_{k-1}\le f(A(k))$ while $A(k)-A(k-1)=\int_{k-1}^k f(A(t))\,dt\ge f(A(k))$, so we must already have $a_{k-1}>A(k-1)+f(0)$, contradicting the choice of $k$.

Similarly, let $k$ be the first index for which $a_k<A(k)-f(0)$ (if it exists). Then $A(k-1)=A(k)-\int_{k-1}^k f(A(t))\,dt\ge A(k)-f(0)>a_k$, so $A(k)-A(k-1)=\int_{k-1}^k f(A(t))\,dt\le f(a_k)$ while $a_k-a_{k-1}=f(a_{k-1})\ge f(a_k)$, so we must already have $a_{k-1}<A(k-1)-f(0)$, contradicting the choice of $k$.

Now just solve the related differential equation and use the fact that $\log(1-1/n)\asymp -1/n$ for large $n$.