Suma: $\sum_{n=1}^{\infty}\ln\left(1+\frac{8}{n^2+9n}\right)$

Question

Tengo un problema con esta suma, llevo un tiempo intentando resolverlo sin éxito.

$$S=\sum_{n=1}^{\infty}\ln\left(1+\frac{8}{n^2+9n}\right)$$

Intenté esto que encontré en mi libro de texto que parece relativo, pero no estoy seguro de cómo aplicarlo al problema: $$a_n=f(n)-f(n+k) \ ,\forall n\geq1$$

Answer 1

Una pista:

$$\ln\left(1+\frac{8}{n^2+9n}\right)=\ln\left(\frac{n^2+9n+8}{n^2+9n}\right)=\ln\left({n^2+9n+8}\right)-\ln\left({n^2+9n}\right)$$

Answer 2

Finalmente,

\begin{eqnarray} &&\sum\limits_{n = 1}^\infty {\ln \left( {1 + \frac{8}{{{n^2} + 9n}}} \right)} \\ &=& \sum\limits_{n = 1}^\infty {\ln \frac{{n + 1}}{n}} - \sum\limits_{n = 1}^\infty {\ln \frac{{n + 8}}{{n + 9}}} \\ &=& \ln \prod\limits_{n = 1}^\infty {\frac{{n + 1}}{n}} - \ln \prod\limits_{n = 1}^\infty {\frac{{n + 8}}{{n + 9}}} \\ &=& \ln \prod\limits_{n = 1}^8 {\frac{{n + 1}}{n}} + \ln \prod\limits_{n = 9}^\infty {\frac{{n + 1}}{n}} - \ln \prod\limits_{n = 1}^\infty {\frac{{n + 8}}{{n + 9}}} \\ &=& \ln \prod\limits_{n = 1}^8 {\frac{{n + 1}}{n}} + \ln \prod\limits_{n = 1}^\infty {\frac{{n + 9}}{{n + 8}}} - \ln \prod\limits_{n = 1}^\infty {\frac{{n + 8}}{{n + 9}}} \\ &=& \ln \prod\limits_{n = 1}^8 {\frac{{n + 1}}{n}} \\ &=& 2\ln 3 \end{eqnarray}

Answer 3

$$a_n =\ln \left(1+\dfrac8{n^2+9n}\right) = \ln \left(\dfrac{n^2+9n+8}{n^2+9n}\right) = \ln (n+1) - \ln(n) + \ln(n+8) - \ln(n+9)$$ Por lo tanto, $$a_n = b_{n+1} - b_n$$ donde $$b_n = \ln(n) - \ln(n+8)$$

Answer 4

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \begin{align} \color{#66f}{\large S}&=\sum_{n = 1}^{\infty}\ln\pars{1 + {8 \over n^2+9n}} =8\sum_{n = 1}^{\infty}\int_{0}^{1}{\dd x \over n^{2} + 9n + 8x} =8\int_{0}^{1}\sum_{n = 1}^{\infty}{1 \over n^{2} + 9n + 8x}\,\dd x \end{align}

La suma sobre $\ds{n}$ puede ser realizado por " estándar significa " . Tras la integración, el resultado es $\begin{array}{|c|}\hline\color{#66f}{\large 2\ln\pars{3} \approx 2.1972}\\ \hline\end{array}$ .