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| − | \textbf{PART A)}\\
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| − | \textbf{To show:}\\
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| − | $ f_{k}(x) \rightarrow f(x) $ for k $ \rightarrow \infty$\\
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| − | \textbf{Proof:}\\
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| − | Take any x $ \in $ (0 ; $\infty$) and $ k > x $ ; consider:
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| − | $ \ln((1-\frac{x}{k})^{k}) = k\ln(1-\frac{x}{k}) = \frac{x}{x}k(\ln(1-\frac{x}{k})-\ln(1)) = -x\dfrac{(\ln(1)-\ln(1-\frac{x}{k}))}{\frac{x}{k}} $\\
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| − | The fraction is the difference quotient of ln evaluated in 1, thus:\\
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| − | $\Rightarrow \lim_{k \to \infty} \ln((1-\frac{x}{k})^{k}) = -x\dfrac{d}{dx}\ln(x)\mid_{x=1} = -x $ \\
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| − | $\Rightarrow \lim_{k \to \infty} f_{k}(x) = x^{t-1} \lim_{k \to \infty} \exp ( \ln((1-\frac{x}{k})^{k}) $
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| − | $ = x^{t-1} \exp(\lim_{k \to \infty} ln((1-\frac{x}{k})^{k}) = x^{t-1} e^{-x} = f(x)$\\
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| − | \textbf{Remark:} since exp is continuous, we can "bring the limes inside" $ \blacksquare $ \\
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| − | \textbf{To show:}\\
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| − | $ f_{k}(x) \leq f_{k+1}(x)$ $\forall x > 0$ \\
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| − | \textbf{Proof:}\\
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| − | \textbf{Case 1:} $ \forall x \in $ $ ]0;k[ $\\
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| − | The arithmetic-geometric mean equality says:\\
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| − | $ \forall x_{1},...,x_{n} \geq 0 $ it holds : $ ( x_{1}...x_{n})^{\frac{1}{n}} \leq \dfrac{x_{1}+...+x_{n}}{n}$\\
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| − | Consider:\\
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| − | $ ((1-\dfrac{x}{k})^{k}\cdot1)^{\frac{1}{k+1}} \leq^{AM-GM} \dfrac{1}{k+1}(k(1-\dfrac{x}{k})+1) = \dfrac{1}{k+1}(k + 1 - x) = \\
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| − | = (1 - \dfrac{x}{k+1}) $\\
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| − | $\Rightarrow (1-\dfrac{x}{k})^{k} \leq (1 - \dfrac{x}{k+1})^{k+1} $\\
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| − | \textbf{Remark:} we don't have any problem of sign or multivalue with the k+1-root because $ (1-\dfrac{x}{k})^{k} $ takes only real positive value, since $ 0 < x < k $.\\
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| − | \textbf{Case 2:} $ x \in [k;k+1[$\\
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| − | The inequality holds since $ f_{k}(x) = 0 $ and $ f_{k+1}(x) = x^{t-1}(1 - \dfrac{x}{k+1})^{k+1} $ is a positve function $ \forall x \in [k;k+1[$.\\
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| − | \textbf{Case 3:} $ x \in [k+1;\infty[$\\
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| − | Both the functions are equal to zero so the inequality still holds.\\
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| − | \newpage
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| − | \textbf{PART B)}\\
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| − | \textbf{Idea:}
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| − | in a first step we show that $I_{k}$ is equal to the right hand side of the equality. Then we rewrite the given integral in a much more convenient way, in particular we want something that looks like the given functions in part a). Eventually we let $k\rightarrow\infty$ and apply Lebesgue's monoton convergence theorem.\\
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| − | $I_{k} = k^{t} \int_0^1 \! u^{t-1}(1-u)^{k} \, \mathrm{d}u = ^{P.I.}\\
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| − | =k^{t}*\lbrace\dfrac{u^{t}}{t}(1-u)^{k}\vert_{0}^{1} - \int_0^1 \! \dfrac{u^{t}}{t}(-1)k(1-u)^{k-1} \, \mathrm{d}u \rbrace = \\
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| − | = k^{t}*\lbrace 0 + \dfrac{k}{t}\int_0^1 \! u^{t}(1-u)^{k-1} \, \mathrm{d}u \rbrace =\\
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| − | = \dfrac{k^{t+1}}{t}\int_0^1 \! u^{t}(1-u)^{k-1} \, \mathrm{d}u =^{P.I.}\\
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| − | =\dfrac{k^{t+1}}{t}*\lbrace\dfrac{u^{t+1}}{t+1}(1-u)^{k-1}\vert_{0}^{1} - \int_0^1 \! \dfrac{u^{t+1}}{t+1}(-1)(k-1)(1-u)^{k-2} \, \mathrm{d}u \rbrace = \\
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| − | = \dfrac{k^{t+1}}{t}*\lbrace 0 + \dfrac{k-1}{t+1}\int_0^1 \! u^{t+1}(1-u)^{k-2} \, \mathrm{d}u \rbrace =\dfrac{k^{t+1}(k-1)}{t(t+1)} \int_0^1 \! u^{t+1}(1-u)^{k-2} \, \mathrm{d}u =^{P.I.}\\
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| − | = ... = \frac{(k-1)!k^{t+1}}{t(t+1)...(t+k-1)} \int_0^1 \! u^{t+k-1} \, \mathrm{d}u = \frac{(k-1)!k^{t+1}}{t(t+1)...(t+k-1)}\dfrac{u^{t+k}}{t+k}\vert^{1}_{0} =$ $\frac{k!k^{t}}{t(t+1)...(t+k)} $ \\
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| − | Where we have used k-1 times partial integration to reach the last but one step.\\
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| − | $ I_{k} = k^{t} \int_0^1 \! u^{t-1}(1-u)^{k} \, \mathrm{d}u = k^{t-1} \int_0^k \! (\dfrac{x}{k})^{t-1}(1-\dfrac{x}{k})^{k} \, \mathrm{d}x = \\
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| − | = \int_0^k \! x^{t-1}(1-\dfrac{x}{k})^{k} \, \mathrm{d}x = \int_0^k \! x^{t-1}(1-\dfrac{x}{k})^{k} \, \mathrm{d}x + \int_k^\infty \! 0 \, \mathrm{d}x = \int_0^k \! f_{k}(x) \, \mathrm{d}x + \int_k^\infty \! f_{k}(x) \, \mathrm{d}x =\\
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| − | = \int_0^\infty \! f_{k}(x) \, \mathrm{d}x $ \\
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| − | where the first step is a change of variable u $=$ $\dfrac{x}{k} $\\
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| − | Letting $ k \rightarrow\infty $, using convergence of exercise 2 a) and Lebesgue's monoton convergence theorem on $ f_{k}(x) $ we obtain:\\
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| − | $ I_{\infty} = \int_0^\infty \! x^{t-1}e^{-x}\, \mathrm{d}x = \Gamma(t)$\\
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| − | Here the conditions to apply Lebesgue's monoton convergence theorem are satisfied, because $ f_{k}(x) $ are positive functions and because $ f_{k}(x) $ is a continuous (and therefore measurable) function on $ [0;\infty[ $.
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| − | \end{document}
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