IIT JEE , AIEEE Forum - Mathematics , Algebra

sandeepramesh

Let $p > 3$ be an odd prime . Show that the numerator of $\sum_{k = 1}^{p - 1} \frac {1}{k}$ is divisible by $p^2$ Smile

computer001

wth do u mean by numerator is divisible by p^2??
i can jus multiply n devide the whole thing by p^2...
so numerator now will be x(p^2) hence it is divisible by p^2..
il also multiply n devide values so tht x is an integer.. then numerator will b div by p^2 no??

sandeepramesh

i mean in the simplest form of the sum as a fraction da :)

as usual simply asking doubts :D :D

computer001

EDITED:
oops..i made a horrible mistake

computer001

EDITED

sandeepramesh

What? Mind posting the soln?

Dei u cant find sum of n terms of a HP

anandghegde

computer 001 i think your harmonic mean is wrong....

$\frac{1}{H} = \frac{\frac{1}{1} +\frac{1}{2}+...+\frac{1}{p-1}}{p-1}$

computer001

no..no i made a mistake in wat i was inverting !
EDITED..
@ mad: yeah thts obvious..i initially got 2 tht but then i made a mistake in the h.p itself..which ob caused me a few moments of hyper-smelling

elastiboysai

sum of p odd nos is p^2.
seper the thing int o fractions wid even n odd denoms.
in the odd thing num will contain p-1 odd nos p-1 times
in even thing take 2 out common u ll get p-1/2 odd nos n again even nos p-1/2. keep writin in terms of odd nos. (may sound long) bt once u get a genl sequence keep doin dis n ull get fract in wich
num will contain odd nos only till p.

sandeepramesh

a soln rather than a hint will be apt for this sum :D bcos its a good one n not so obvious :)

elasti, how can yuo be soo sure that there will be no cancellations between the nr n denominator?

konichiwa2x

edited. check my post below for a better explanation.

sandeepramesh

Can you be more clear abt the start? And does already p divide B? What B have you set up?

Decoder

my steps same till sum{ k^-2}....uske baad i culdn't get further..
can u explain wat u did kon!

konichiwa2x

This question is an application of wolstenholem'es theorem. Let

we want to prove that

divides

we have

.

Ttherefore:

So we need to prove that

divides

.

By Wilson's theorem,

now since

, it's in the form

In either case

will divide

thus

divides

and we're done!

sandeepramesh

nice :)

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