근의 제곱합 극한

$\displaystyle \mathrm{S} _{n} = \sum\limits _{{k} = 1} ^{n} \left( \frac{1}{2} \right) ^{n-1} = \frac{1 - \left( \frac{1}{2} \right) ^{n}}{1 - \frac{1}{2}} = 2 - \left( \frac{1}{2} \right) ^{n-1}$ $\displaystyle T _{n} = \sum\limits _{k = 1} ^{n} \frac{1}{k ( k + 1 )} = \sum\limits _{k = 1} ^{n} \left( \frac{1}{k} - \frac{1}{k + 1} \right) = 1 - \frac{1}{n + 1}$ 이차방정식 $x ^{2} - S _{n} x + T _{n} = 0$의 두 근이 $\alpha _{n} , \beta _{n}$이므로 근과 계수와의 관계에서 $\alpha _{n} + \beta _{n} = \mathrm{S} _{n} , \alpha _{n} \beta _{n} = \mathrm{T} _{n}$ $\begin{aligned} \alpha _{n} ^{2} + \beta _{n} ^{2} & = \left( \alpha _{n} + \beta _{n} \right) ^{2} - 2 \alpha _{n} \beta _{n} = \mathrm{S} _{n} \mathit{^{2}} - 2 \mathrm{T} _{n} \end{aligned}$$\displaystyle = \left\{ 2 - \left( \frac{1}{2} \right) ^{n - 1} \right\} ^{2} - 2 \times \left( 1 - \frac{1}{n + 1} \right)$$\displaystyle = 2 - 4 \times \left( \frac{1}{2} \right) ^{n - 1} + \left( \frac{1}{4} \right) ^{n - 1} + 2 \times \frac{1}{n + 1}$ $\displaystyle \therefore \lim\limits _{n \rightarrow \infty} \left( \alpha _{n} ^{2} + \beta _{n} ^{2} \right) = 2$