The honey jar

A jar holds 640 grams of honey. A wise bear decides: every day I take half of what is left in the jar.

Jar: $J_0=640$ g. Each day the bear takes half of what is left.

$J_0=640$, $J_k=J_{k-1}/2$

On day one he takes half of 640, which is 320 grams, and 320 grams are left. On day two he takes 160 and 160 are left. On day three, 80. Each day the jar holds half of what it held the day before.

On day $k$ the bear takes $\frac{640}{2^k}$ g and as much is left. Days $1,2,3$: $320,160,80$ g.

$(J_1,J_2,J_3)=(320,160,80)$

After seven days 5 grams are left (640 divided by 128), and after ten days 0.625 grams, less than one gram. But the jar never becomes empty: after any number of days, something is in it.

After $n$ days there is left $J_n=640\cdot 2^{-n}$. We have $J_7=5$ and $J_{10}=0.625$, and for every $n$ we have $J_n>0$.

$J_n=640\cdot 2^{-n}$, $J_7=5$, $J_{10}=0.625$, $\forall n,J_n>0$

Even so, the bear has eaten almost the whole jar: after ten days, 639.375 grams. If the days went on without end, the total eaten would come closer and closer to 640, but never reach it. There is always honey for tomorrow; enough means taking only a part, each time.

Eaten after $n$ days: ${\sum }_{k=1}^n\frac{640}{2^k}=640(1-2^{-n})$; for $n=10$ it is $639.375$. It gets closer to $640$ as $n$ grows: ${\lim }_{n\to \infty }640(1-2^{-n})=640$, but never reaches it.

${\sum }_{k=1}^n\frac{640}{2^k}=640(1-2^{-n})<640$, ${\lim }_{n\to \infty }640(1-2^{-n})=640$