matematikk.net

finn verdien av følgende sum:

[tex]\sum_{n=1}^\infty\frac{1}{n}\left(\frac{1}{n+1}\,-\,\frac{1}{n+2}\,+\,\frac{1}{n+3}\,-\,...\right)[/tex]

[tex]\sum_{n=1}^{\infty}\sum_{m=1}^{\infty} \frac{(-1)^{m+1}}{n(n+m)} [/tex]

Legg merke til at [tex]\int_0^{1} r^{n+m-1}\,dr=\frac{1}{n+m}[/tex].

Ved å anta at vi kan gjøre ombytte av sum og integral fås

[tex]-\int_0^1 dr \frac{1}{r}\left [\sum_{n=1}^{\infty} \frac{r^n}{n}\right ]\left [\sum_{m=1}^{\infty}(-r)^m\right ][/tex]

Igjen er [tex]\frac{r^n}{n} =\int_0^r ds\, s^{n-1}[/tex], så ved å bytte om på summen fås

[tex]-\int_0^1 dr \frac{1}{r}\left [\int_0^r ds\,\frac{1}{s}\sum_{n=1}^{\infty} s^n\right ]\left [\sum_{m=1}^{\infty}(-r)^m\right ][/tex]

Ved å bruke formler for geometriske rekker fås

[tex]-\int_0^1 dr \frac{\ln(1-r)}{1+r}\approx 0.582241...[/tex]

så vidt jeg henger med i svingene stemmer dette...

[tex]\text eksakt sum:\,\,{1\over 2}\left(\zeta(2)-\ln^2(2)\right)=\frac{\pi^2}{12}\,-\,\frac{\ln^2(2)}{2}={1\over 2}\left(Li_2(1)-\ln^2(2)\right)[/tex]

Hvordan løste du den selv?

plutarco skrev:Hvordan løste du den selv?

ikke helt aleine, fikk litt hjelp av en matematiker...
skal skrable ned noe seinere...

[tex]J=\sum_{n=1}^\infty\frac{1}{n}\left(\frac{1}{n+1}\,-\,\frac{1}{n+2}\,+\,\frac{1}{n+3}\,-\,...\right)[/tex]
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har at:
[tex]\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+3}-...=\int_0^1\left(x^n-x^{n+1}+x^{n+2}-...\right)\,dx=\int_0^1\frac{x^n}{x+1}\,dx[/tex]
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[tex]J=\sum_{n=1}^{\infty}{1\over n}\int_0^1\frac{x^n}{x+1}\,dx=\int_0^1\frac{1}{x+1}\left(\sum_{n=1}^{\infty}{x^n\over n}\right)\,dx=-\int_0^1\frac{\ln(1-x)}{x+1}\,dx=-\int_0^1\frac{\ln(y)}{2-y}\,dy=-\int_0^1\frac{\ln(y/2)+\ln(2)}{2-y}\,dy=-\int_0^1\frac{\ln(y/2)}{2-y}\,dy\,-\,\ln^2(2)=-\int_0^{1/2}\frac{\ln(u)}{1-u}\,du\,-\,\ln^2(2)=-\int_{1/2}^1\frac{\ln(1-u)}{u}\,du\,-\,\ln^2(2)[/tex]

bruker substitusjonene: y = x - 1 og u = y/2 underveis...

[tex]J=-\left(\int_0^1\,-\,\int_0^{1/2}\right)\frac{\ln(1-u)}{u}\,du-\ln^2(2)={1\over 2}\left(\zeta(2)-\ln^2(2)\right)[/tex]

Jeg la inn align for deg:

[tex]\begin{align} J = \sum_{n=1}^{\infty}{1\over n}\int_0^1\frac{x^n}{x+1}\,dx &= \int_0^1\frac{1}{x+1}\left(\sum_{n=1}^{\infty}{x^n\over n}\right)\,dx \\ &= -\int_0^1\frac{\ln(1-x)}{x+1}\,dx \\ &=-\int_0^1\frac{\ln(y)}{2-y}\,dy \\ &= -\int_0^1\frac{\ln(y/2)+\ln(2)}{2-y}\,dy \\ &=-\int_0^1\frac{\ln(y/2)}{2-y}\,dy\,-\,\ln^2(2) \\ &=-\int_0^{1/2}\frac{\ln(u)}{1-u}\,du\,-\,\ln^2(2) \\ &= -\int_{1/2}^1\frac{\ln(1-u)}{u}\,du\,-\,\ln^2(2) \end{align}[/tex]

ja, det var mer oversiktlig...