# elem#theo - induction + Bertrand's

• Oct 22nd 2008, 02:33 PM
cassiopeia1289
elem#theo - induction + Bertrand's
OK: $\displaystyle p_{n}$ denotes the $\displaystyle nth$ prime. For $\displaystyle n>3$, show that $\displaystyle p_{n} < p_{1} + p_{2} + ... + p_{n-1}$
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, $\displaystyle p_{4} = 7 < 2+3+5$, which is true.
Assume for some k, $\displaystyle p_{k} < p_{1} + p_{2} + ... + p_{k-1}$
By Bertrand, there exists a prime between $\displaystyle p_{k}$ and $\displaystyle 2p_{k}$

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

How? *so confused*
• Oct 23rd 2008, 11:30 AM
cassiopeia1289
anyone? please - kinda urgent ...
• Oct 23rd 2008, 01:44 PM
cassiopeia1289
HELP!!!
• Oct 23rd 2008, 02:12 PM
chiph588@
by our inductive assumption, $\displaystyle p_{k} < p_{1} + ... + p_{k-1}$

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

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

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

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

$\displaystyle p_{k+1} < p_{1} + ... + p_{k}$
• Oct 23rd 2008, 04:04 PM
cassiopeia1289
Ah - thank you!