Question

va rog sa ma ajutati sa demonstrez asta​

Answer 1

$S = \dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{10000}}$

$\displaystyle 198 <\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{10000}}<199\\ \\\\\text{Observam ca functia }\dfrac{1}{\sqrt{x}} \text{ e strict descrescatoare} \\ \text{cand }x>0, \text{ astfel putem deduce ca pentru }x\in [k,k+1]:\\ \\\dfrac{1}{\sqrt{x+1}}\leq \dfrac{1}{\sqrt{k+1}}<\dfrac{1}{\sqrt{x}}\Bigg|\int_{k}^{k+1}\, dx\\ \\ \Leftrightarrow \int_{k}^{k+1}\dfrac{1}{\sqrt{x+1}}\, dx\leq \int_{k}^{k+1}\dfrac{1}{\sqrt{k+1}}\, dx<\int_{k}^{k+1}\dfrac{1}{\sqrt{x}}$

$\displaystyle \Leftrightarrow \int_{k}^{k+1}\dfrac{1}{\sqrt{x+1}}\, dx\leq \dfrac{k+1-k}{\sqrt{k+1}}\, dx<\int_{k}^{k+1}\dfrac{1}{\sqrt{x}}\\ \\ \Leftrightarrow \int_{k}^{k+1}\dfrac{1}{\sqrt{x+1}}\, dx\leq \dfrac{1}{\sqrt{k+1}}\, dx<\int_{k}^{k+1}\dfrac{1}{\sqrt{x}}\\ \\\\\text{Adunam inegalitatile pentru }k = 1,2,3,...,9999.\\\\ \\ \Leftrightarrow \int_{1}^{10000} \dfrac{1}{\sqrt{x+1}}\,dx \leq\sum\limits_{k=1}^{9999}\dfrac{1}{\sqrt{k+1}}< \int_{1}^{10000} \dfrac{1}{\sqrt{x}}\, dx$

$\displaystyle \Leftrightarrow \dfrac{(x+1)^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\Bigg|_{1}^{10000} \leq \sum\limits_{k=1}^{9999}\dfrac{1}{\sqrt{k+1}}< \dfrac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\Bigg|_{1}^{10000}\\ \\ \Leftrightarrow 2\sqrt{10001}-2\sqrt{2}\leq S -1 < 2\sqrt{10000}-2 \\ \\ \Leftrightarrow 2\sqrt{10001}-2\sqrt{2}\leq S -1 < 198 \Big|+1 \\ \\ \Leftrightarrow 2\sqrt{10001}-2\sqrt{2}+1\leq S< 199 \\ \\ \Leftrightarrow 198 < 2\sqrt{10001}-2\sqrt{2}+1\leq S< 199$

⇒ 198 < S < 199 (q.e.d.)