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I proved that the serie $\displaystyle{\sum_{n = 1}^{\infty}{n \over 3^{n}}}$ is convergent, but I want to find the value of the sum using the Cauchy product. Any suggestion ?.

Felix Marin
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coolsv
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5 Answers5

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Let $$a_n=\frac 1{3^{n}} \qquad b_n=\frac 1{3^{n}}$$ then $$\left(\sum_{i=0}^\infty \frac 1{3^{n}}\right)\times \left(\sum_{i=0}^\infty \frac 1{3^{n}}\right)=\sum_k\sum_{l=0}^k \frac 1{3^{l}}\times \frac{3^l}{3^k}$$

$$\left(\sum_{i=0}^\infty \frac 1{3^{n}}\right)^2=\sum_k\frac 1{3^k}\times(k+1)$$ $$ \left(\sum_{i=0}^\infty \frac 1{3^{n}}\right)^2=x+\left(\sum_{i=0}^\infty \frac 1{3^{n}}\right)$$ Where $$x=\sum_k \frac{k}{3^k}$$ Geometric series (can't we use this particular case at least?): $$\left(\sum_{i=0}^\infty \frac 1{3^{n}}\right)=\frac 32$$

Hence $$\frac 94=\frac 32 +x$$

Finally $$x=\frac 34$$

entrelac
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You don't need the Cauchy product. It is sufficient to note that $$\sum_{n\geq0}x^{n}=\frac{1}{1-x},\,\left|x\right|<1 $$ hence taking the derivative $$\sum_{n\geq0}nx^{n-1}=\frac{1}{\left(1-x\right)^{2}} $$ so $$\sum_{n\geq0}nx^{n}=\sum_{n\geq1}nx^{n}=\frac{x}{\left(1-x\right)^{2}} $$ hence taking $x=\frac{1}{3} $ we have $$\sum_{n\geq1}\frac{n}{3^{n}}=\color{red}{\frac{3}{4}}.$$

quid
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Marco Cantarini
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    Yes that's perfect, but the intent of the question is to use the Cauchy product, as we don't know the first result (the power series) and the use of the derivaties :-) – coolsv Jul 17 '16 at 19:41
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Use $\sum_{n\geq 0}x^{n}=\frac{1}{1-x}$ for $|x|<1$. \begin{align} \frac{1}{(1-x)^{2}}=\left(\sum_{n\geq 0} x^{n}\right)^{2}=\sum_{m,n\geq0}x^{m+n}=\sum_{r\geq0}(r+1)x^{r} \end{align} (Here we use Cauchy product.) Then we have \begin{align} \sum_{r\geq0}rx^{r}=\sum_{r\geq0}(r+1)x^{r}-\sum_{r\geq0}x^{r}=\frac{1}{(1-x)^{2}}-\frac{1}{1-x} \end{align}

Seewoo Lee
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  • Thank you! Good approach too!! But as I said, we are not supposed to use the result of the power series. ;-) – coolsv Jul 17 '16 at 20:20
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$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \color{#f00}{\sum_{n = 1}^{\infty}{n \over 3^{n}}} & = \sum_{n = 1}^{\infty}{n \choose n - 1}\pars{1 \over 3}^{n} = {1 \over 3}\sum_{n = 0}^{\infty}{n + 1 \choose n}\pars{1 \over 3}^{n} \\[4mm] & = {1 \over 3}\sum_{n = 0}^{\infty} {-n - 1 -n - 1 \choose n}\pars{-1}^{n}\pars{1 \over 3}^{n} = {1 \over 3}\sum_{n = 0}^{\infty}{-2 \choose n}\pars{-\,{1 \over 3}}^{n} \\[4mm] & = {1 \over 3}\bracks{1 + \pars{-\,{1 \over 3}}}^{-2} = \color{#f00}{3 \over 4} \end{align}

Felix Marin
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Another approach that uses the geometric series is to recognize that

$$\begin{align} \sum_{n=1}^N \frac{n}{3^n}&=\sum_{n=1}^N \frac{1}{3^n}\sum_{m=1}^n(1) \end{align}$$

Now, interchanging the order of summation reveals

$$\begin{align} \sum_{n=1}^N \frac{n}{3^n}&=\sum_{m=1}^N \sum_{n=m}^N\frac{1}{3^n}\\\\ &=\sum_{m=1}^N \frac{(1/3)^m-(1/3)^{N+1}}{1-1/3}\\\\ &=\frac32 \left(\frac{(1/3)-(1/3)^{N+1}}{1-1/3}-N(1/3)^{N+1}\right)\\\\ &\to \frac34\,\,\text{as}\,\,N\to \infty \end{align}$$

Mark Viola
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