Question

$\text{Fie sirul }(x_n)_{n\geq 1},\text{dephinit prin } x_n=\displaystyle \sum_{k=1}^n \dfrac{1}{2^{k^2}}.\text{Aratati ca limita}\\ \text{sirului este un numar irational.}$

Answer 1

$\displaystyle Am~mai~vazut~si~eu~ceva~asemanator~candva,~dar~cu~3~in~loc \\ \\ de~2.~(Gheorghe~Lazar,~2003) \\ \\ Convergenta~se~demonstreaza~usor: ~Sirul~este~strict~crescator~si \\ \\ marginit~superior~de~2. \\ \\ \left(pentru~ca~x_n \le \frac{1}{2}+ \frac{1}{2^2}+...+ \frac{1}{2^n}\ \textless \ 2 \right) \\ \\ In~fine...~numarul~dat~este~de~fapt~un~numar \\ \\ scris~in~baza~2,~si~anume: \\ \\ 0,100100001000... \\ \\ Cifrele~de~1~ocupa~pozitiile~1,4,9,16,...~numaratoarea~incepand~cu \\ \\ prima~zecimala.$

$\displaystyle Se~observa~ca~intre~cifrele~consecutive~de~1~se~afla \\ \\ 2,4,6,8...~cifre~de~0.~(in~ordinea~aceasta) \\ \\ Deci~este~clar~ca~numarul~scris~in~baza~2~nu~este~periodic. \\ \\ Iar~pentru~ca~are~o~infinitate~de~cifre,~rezulta~ca~este~irational. \\ \\ Acest~ultim~fapt~implica^*~irationalitatea~numarului~in~baza~10. \\ \\ ----------------\\ \\^*voi~incerca~sa~explic~in~comentariu...~in~esenta~nu~e~greu \\ \\ de~inteles,~dar~e~greu~de~explicat.$