# Thread: elem#theo - induction + Bertrand's

1. ## elem#theo - induction + Bertrand's

OK: $p_{n}$ denotes the $nth$ prime. For $n>3$, show that $p_{n} < p_{1} + p_{2} + ... + p_{n-1}$
the hint was to use induction and Bertrand's conjecture

ok: so the answer is in the book, but its not helpful b/c I don't get one jump

so they do this:
So for n=4, $p_{4} = 7 < 2+3+5$, which is true.
Assume for some k, $p_{k} < p_{1} + p_{2} + ... + p_{k-1}$
By Bertrand, there exists a prime between $p_{k}$ and $2p_{k}$

Here is where I'm lost: they go from there to $p_{k+1} \leq 2p_{k} < (p_{1} + p_{2} + ... + p_{k_{1}})$,
thus proving that statement is true for $k+1$

How? *so confused*

2. anyone? please - kinda urgent ...

3. HELP!!!

4. by our inductive assumption, $p_{k} < p_{1} + ... + p_{k-1}$

so add $p_{k}$ to both sides:

$p_{k} + p_{k} < p_{1} + ... + p_{k}$

which is the same thing as $2p_{k} < p_{1} + ... + p_{k}$

and we know $p_{k+1} \leq 2p_{k}$ by the conjecture, so...

$p_{k+1} < p_{1} + ... + p_{k}$

5. Ah - thank you!