Ant climbing on bush

For simplicity, let $0.1m$ be the unit. Assume that on night $n$ the ant is at height $a_n$ and the bush has height $b_n$ (all measured in units, $a_0=0$, $b_0=100$). So the new height of the bush is $b_{n+1}=b_n+5$, which yields the arithmetic series $b_n=5n+100$. The new position of the ant is $a_{n+1}= (a_n+1)\cdot \frac{b_n+5}{b_n}= (a_n+1)\cdot \frac{n+21}{n+20}$. (Easy calculation.)

As an auxiliary series, we introduce $x_n:= \frac{a_n}{n+20}$. Then the above formula translates to $c_{n+1}= c_n+\frac{1}{n+20}$.

Thus $c_n$ is almost the harmonic series. (The harmonic series is $H_n= \sum\limits_{k=1}^n \frac{1}{k}$. It is well-known that $H_n>\log n$.)

In fact $c_n= H_{n+20}-H_{20}$, and in particular $a_n=(H_{n+20}-H_{20})(n+20)$, which beats $b_n=5(n+20)$ as soon as $H_{n+20}-H_{20}\geq 5$.

With precise calculation, this yields that the ant reaches the top exactly on day $3043$, so you must have made a rounding error with Excel.

The total height of the bush after $d$ days is $h_d=(10 + 0.5d){\text{m}}$. When the ant climbs $0.1{\text{m}}$ at night, this constitutes a fraction $0.1 / h_d = \frac{1}{100 + 5d}$ of the total height. (Of course, when the bush grows during the day, the ant's relative position on the bush is not affected.) So you just need to know for what value of $N$ the sum $\sum_{d=1}^{N}\frac{1}{100+5d}$ first exceeds $1$.

This is if the bush grows first, before the ant ever climbs... your original formulation gives the ant a head start relative to this, because you let the ant climb before the bush grows. In that case, you'd want to know when $\sum_{d=1}^{N}\frac{1}{100+5d}$ exceeds $0.99$. Using Excel, I find that the climb takes $2874$ days for the latter case (in agreement with OP) and $3023$ days for the former case.