Valutazione di un limite senza utilizzo della somma di Riemann
Valutazione di$$\lim_{n \rightarrow \infty}\bigg[\frac{1}{n}+\frac{1}{n+2}+\frac{1}{n+4}+\cdots \cdots +\frac{1}{3n}\bigg]$$
Il mio lavoro: usare la somma di Riemann
$$\lim_{n\rightarrow \infty}\sum^{n}_{r=0}\frac{1}{n+2r}=\lim_{n\rightarrow\infty}\sum^{n}_{r=0}\frac{1}{1+2\frac{r}{n}}\cdot \frac{1}{n}$$
Mettere$\displaystyle \frac{r}{n}=x$e$\displaystyle \frac{1}{n}=dx$e modificando i limiti
$$\int^{1}_{0}\frac{1}{1+2x}dx=\frac{1}{2}\ln|1+2x|\bigg|^{1}_{0}=\frac{1}{2}\ln(3)$$
Ma sarebbe possibile risolvere il problema senza utilizzare la somma di Rienmann? come in, si potrebbero risolvere tali somme infinite in un metodo alternativo alla conversione della somma in un integrale.
Risposte
Per pari$n=2m$noi abbiamo\begin{align} \sum_{r=0}^{2m}\frac{1}{2m+2r} &=\frac{1}{2}\sum_{r=0}^{2m}\frac{1}{m+r}=\\ &=\frac{1}{2}\left(\sum_{r=1}^{3m}\frac{1}{r}-\sum_{r=1}^{m-1}\frac{1}{r}\right)=\frac{1}{2}(H_{3m}-H_{m-1}), \end{align}dove$$ H_n=\sum_{r=1}^n\frac{1}{r} $$sono i numeri armonici . Data la nota relazione$$ \lim_{n\to\infty}(H_n-\log n)=\gamma $$noi abbiamo\begin{align} &\lim_{m\to\infty}\sum_{r=0}^{2m}\frac{1}{2m+2r} =\frac{1}{2}\lim_{m\to\infty}(H_{3m}-H_{m-1})=\\ &\qquad=\frac{1}{2}\lim_{m\to\infty}[(H_{3m}-\log(3m))+\log(3m)-(H_{m-1}-\log(m-1))-\log(m-1)]=\\ &\qquad=\frac{1}{2}\lim_{m\to\infty}[\gamma+\log(3m)-\gamma-\log(m-1)]=\\ &\qquad=\frac{1}{2}\lim_{m\to\infty}\log\left(\frac{3m}{m-1}\right)=\frac{1}{2}\log 3. \end{align}
Per dispari$n=2m+1$, tenere in considerazione\begin{align} &\frac{1}{n}+\frac{1}{n+2}+\ldots+\frac{1}{3n-2}+\frac{1}{3n}=\\ &=\left(\frac{1}{n}+\frac{1}{n+1}+\ldots+\frac{1}{3n-1}+\frac{1}{3n}\right)-\left(\frac{1}{n+1}+\frac{1}{n+3}+\ldots+\frac{1}{3n-3}+\frac{1}{3n-1}\right) \end{align}possiamo scrivere\begin{align} \sum_{r=0}^{2m+1}\frac{1}{2m+1+2r} &= \sum_{s=0}^{4m+2}\frac{1}{2m+1+s}-\sum_{r=0}^{2m}\frac{1}{2m+2+2r}=\\ &= \sum_{s=0}^{4m+2}\frac{1}{2m+1+s}-\frac{1}{2}\sum_{r=0}^{2m}\frac{1}{m+1+r}=\\ &= H_{6m+3}-H_{2m}-\frac{1}{2}[H_{3m+1}-H_{m}] \end{align}e\begin{align} \lim_{m\to\infty}\sum_{r=0}^{2m+1}\frac{1}{2m+1+2r} &= \lim_{m\to\infty}\left(H_{6m+3}-H_{2m}-\frac{1}{2}[H_{3m+1}-H_{m}]\right)=\\ &= \lim_{m\to\infty}\left(\log(6m+3)-\log(2m)-\frac{1}{2}[\log(3m+1)-\log(m)]\right)=\\ &= \lim_{m\to\infty}\left(\log\left(\frac{6m+3}{2m}\right)-\frac{1}{2}\log\left(\frac{3m+1}{m}\right)\right)=\frac{1}{2}\log 3 \end{align}
Prova alternativa
Riscriviamo la somma come$$ \frac{1}{2}\sum_{r=0}^n\frac{1}{\frac{n}{2}+r}=\frac{1}{2}\left[\psi\left(\frac{3n+2}{2}\right)-\psi\left(\frac{n}{2}\right)\right], $$dove$\psi$è la funzione digamma e dove abbiamo usato l'equazione alle differenze$$ \psi(x+N)-\psi(x)=\sum_{k=0}^{N-1}\frac{1}{x+k}, $$vedi Digamma::Formula di ricorrenza e caratterizzazione .
Ora, tenendo conto della seguente disuguaglianza, valida per$x>0$ $$ \log x-\frac{1}{x}\leq\psi(x)\leq\log x-\frac{1}{2x}, $$vedi Digamma::Inequalities , abbiamo$$ \log\left(\frac{3n+2}{n}\right)-\frac{2}{3n+2}+\frac{1}{n}\leq\psi\left(\frac{3n+2}{2}\right)-\psi\left(\frac{n}{2}\right)\leq \log\left(\frac{3n+2}{n}\right)-\frac{1}{3n+2}+\frac{2}{n} $$e dal teorema di compressione, otteniamo il risultato.
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\Large\left. a\right)}$ \begin{align} &\bbox[5px,#ffd]{\sum_{r = 0}^{n}{1 \over n + 2r}} = \sum_{r = 0}^{n}\int_{0}^{1}t^{n + 2r - 1}\,\dd t = \int_{0}^{1}\sum_{r = 0}^{n}t^{n + 2r - 1}\,\dd t \\[5mm] = &\ \int_{0}^{1}t^{n - 1}\,{t^{2n + 2} - 1 \over t^{2} - 1}\,\dd t = \int_{0}^{1}{t^{n - 1} - t^{3n + 1} \over 1 - t^{2}}\,\dd t = {1 \over 2}\int_{0}^{1}{t^{n/2 - 1} - t^{3n/2} \over 1 - t}\,\dd t \\[5mm] = &\ {1 \over 2}\pars{\int_{0}^{1}{1 - t^{3n/2} \over 1 - t}\,\dd t - \int_{0}^{1}{1 - t^{n/2 - 1} \over 1 - t}\,\dd t} = {H_{3n/2} - H_{n/2 -1} \over 2} \\[5mm] \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\, & {\bracks{\vphantom{\Large A}\ln\pars{3n/2} + \gamma + 1/\pars{3n}} - \bracks{\vphantom{\Large A}\ln\pars{n/2 - 1} + \gamma + 1/\pars{n - 2}}\over 2} \\[5mm] \stackrel{\mrm{as}\ n\ \to\ \infty}{\Large\to}\,\,\, & \bbx{\ln\pars{3} \over 2} \\ & \end{align}
$\ds{\Large\left. b\right)}$ \begin{align} &\bbox[5px,#ffd]{\sum_{r = 0}^{n}{1 \over n + 2r}} = \sum_{r = 0}^{\infty}\pars{{1 \over n + 2r} - {1 \over 3n + 2 + 2r}} \\[5mm] = & {1 \over 2}\sum_{r = 0}^{\infty}\pars{{1 \over r + n/2} - {1 \over r + 3n/2 + 1}} = \bbx{H_{3n/2} - H_{n/2 - 1} \over 2}, \quad\mbox{See}\ {\Large\left. a\right)}.\\ & \end{align}