Ant climbing on bush
An ant is on the ground and trying to climb on a (straight) ivy bush 10m high. It crawls up 0.1m each night, but at day, the bush grows uniformly by 0.5m (in its entire height). Will the ant ever reach the top of the ivy? If yes, in how many days? If no, justify why.
Let's convert everything in centimeters. I have made a table with the values of the bush height every night and every day, basis the rules. The ant starts with 1000 cm height above and 0 cm below. Then it advances by 10 cm so it has 990 cm above and 10 cm below. Total height is (still) 1000 cm. In the morning, the height of the bush is proportionally extended by 5% so the up value is now $990*(5/1000)+990 = 1039.5$ and the down is $10*(5/1000)+10 = 10.5$ Total height is of course increased by 50 cm. Then at night, first value is decreased by 10 and second is increased by 10. Height remains the same. We continue this way and have the following values:
Bush up (cm) Bush down (cm) Total
Initially 1000 0 1000
1st night 990 10 1000
1st day 1039,5 10,5 1050
2nd night 1029,5 20,5 1050
2nd day 1078,52381 21,47619048 1100
3rd night 1068,52381 31,47619048 1100
3rd day 1117,093074 32,90692641 1150
4th night 1107,093074 42,90692641 1150
4th day 1155,227555 44,77244495 1200
5th night 1145,227555 54,77244495 1200
We continue this way and now we make a graph of the 3 columns, bush up, bush down and Total. We notice that the 3 curves are declining, which means that they will never intersect. This means that the ant will never reach the top. However, this is not correct. I was told that the problem has a positive reply. Where am I wrong?
Thank you very much in anticipation!
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.
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.