Question

$1+\frac{1}{2} +\frac{1}{3} +\frac{1}{4}+...+\frac{1}{n} =S$ demonstrati ca S nu este numar intreg

Answer 1

$\displaystyle\bf\\notam~~M=produsul~tuturor~numerelor~impare~mai~mici~sau~egale~decat~n.\\notam~2^k=cea~mai~mare~putere~a~lui~2,~cu~conditia~ca~2^k\leq n.\\atunci,~daca~inmultim~ambii~membri~ai~relatiei~cu~M\cdot2^k,~obtinem:\\\frac{M\cdot2^k}{1}+\frac{M\cdot2^k}{2}+\frac{M\cdot2^k}{3}+...+\frac{M\cdot2^k}{2^k}+...+\frac{M\cdot2^k}{n}=MS\cdot2^k.\\fie~1\leq i\leq n,~avem~doua~cazuri~in~functie~de~paritatea~lui~i:$

$\displaystyle\bf\\daca~i=par\implies \frac{M\cdot2^k}{i}=par.\\daca~i=impar\implies \frac{M\cdot2^k}{i}=par,~pentru~ca~in~M~se~gaseste~in\\produs~si~i,~intrucat~am~considerat~i\leq n.\\Dar,~in~suma,~avem~ca~singurul~termen~impar~este~\frac{M\cdot2^k}{2^k},\\pentru~ca~se~va~simplifica~2^k~cu~2^k,~si~vom~ramane~cu~numarul~M\\care~este~impar.$

$\displaystyle\bf\\deci,~membrul~stang~este~impar~pentru~ca~avem~un~singur~termen~\\impar~iar~restul~sunt~toti~pari,~iar~membrul~drept~este~par,~si~cum\\un~numar~nu~poate~fi~par~si~impar~simultan,~am~obtinut~o\\contradictie,~deci~S~nu~poate~fi~numar~intreg.$