2; 2½; 2⅓; 2¼... - числовая последовательность, начиная со второго члена, монотонно убывающая.
1 способ.
1,~n\in N}} \right." alt="\displaystyle a_n=\left \{ {{2,~~~~~~~~~~~~n=1;} \atop {2+\dfrac 1n,~~n>1,~n\in N}} \right." align="absmiddle" class="latex-formula">
Проверка :
n = 1; 
n = 2; 
n = 3; 
n = 4; 
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2 способ.
где
[(n-1):n] - математическое округление "до целого". Результат деления (n-1):n будет равен 0 при n=1, и будет округляться к 1 при любых n>1.
Проверка:
n = 1; ![a_1=2+\dfrac{[(1-1):1]}1=2+\dfrac{0}1=2 a_1=2+\dfrac{[(1-1):1]}1=2+\dfrac{0}1=2](https://tex.z-dn.net/?f=a_1%3D2%2B%5Cdfrac%7B%5B%281-1%29%3A1%5D%7D1%3D2%2B%5Cdfrac%7B0%7D1%3D2)
n = 2; ![a_2=2+\dfrac{[(2-1):2]}2=2+\dfrac{[0,5]}2=2+\dfrac{1}2=2\dfrac{1}2 a_2=2+\dfrac{[(2-1):2]}2=2+\dfrac{[0,5]}2=2+\dfrac{1}2=2\dfrac{1}2](https://tex.z-dn.net/?f=a_2%3D2%2B%5Cdfrac%7B%5B%282-1%29%3A2%5D%7D2%3D2%2B%5Cdfrac%7B%5B0%2C5%5D%7D2%3D2%2B%5Cdfrac%7B1%7D2%3D2%5Cdfrac%7B1%7D2)
n = 3; ![a_3=2+\dfrac{[(3-1):3]}3=2+\dfrac{[0,(6)]}3=2+\dfrac{1}3=2\dfrac{1}3 a_3=2+\dfrac{[(3-1):3]}3=2+\dfrac{[0,(6)]}3=2+\dfrac{1}3=2\dfrac{1}3](https://tex.z-dn.net/?f=a_3%3D2%2B%5Cdfrac%7B%5B%283-1%29%3A3%5D%7D3%3D2%2B%5Cdfrac%7B%5B0%2C%286%29%5D%7D3%3D2%2B%5Cdfrac%7B1%7D3%3D2%5Cdfrac%7B1%7D3)
n = 4; ![a_4=2+\dfrac{[(4-1):4]}4=2+\dfrac{[0,75]}4=2+\dfrac{1}4=2\dfrac{1}4 a_4=2+\dfrac{[(4-1):4]}4=2+\dfrac{[0,75]}4=2+\dfrac{1}4=2\dfrac{1}4](https://tex.z-dn.net/?f=a_4%3D2%2B%5Cdfrac%7B%5B%284-1%29%3A4%5D%7D4%3D2%2B%5Cdfrac%7B%5B0%2C75%5D%7D4%3D2%2B%5Cdfrac%7B1%7D4%3D2%5Cdfrac%7B1%7D4)