A natural number k is called stable if there exists k distinct natural numbers $a_1$,$a_2$,$a_3$....$a_k$ > 1 $$\sum_{i=1}^k \frac{1}{a_i} = 1$$ Show that if k is stable then k+1 is also stable . Using this or otherwise find all stable numbers.
I can understand that this will be done by mathematical induction for natural numbers > 3 . But I don't know how to proceed with this sum.. Please help!
On
As you pointed out it is about induction. I'll give you a hint about the induction step.
Assume that $k$ is stable. Then we have $a_1, \ldots a_k$ with $\sum_{i=1}^k \frac{1}{a_i} =1$. Now we can simply change these number a bit, such that we have $\sum_{i=1}^k \frac{1}{a_i} = \frac{1}{2}$. Then we make a "smart" choice of $a_{k+1}$ and have that also $k+1$ is stable. You must add some proof, that the new $a_i$ are all distinct.
If this is not sufficient to help you, please ask.
The induction step could be as follows:
You have for some $k$ $$\sum_{i=1}^k \frac{1}{a_i} = 1$$ so $$\sum_{i=1}^k \frac{1}{2 a_i} = \frac{1}{2}$$ $$\frac{1}{2} + \sum_{i=1}^k \frac{1}{2 a_i} = 1$$ $$\sum_{i=1}^{k+1} \frac{1}{b_i} = 1$$ where $b_i=2 a_i$, $b_{k+1}=2$. Since the $a_i$ are distinct and $>1$, so are the $b_i$, so $k+1$ is indeed stable.