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Prove that $$\prod_{n=1}^{\infty}\left(\frac{2n+3}{2n+2}\right)$$ converges or diverges.

Taking analysis this year has really intrigued me. We do not cover this topic in analysis, but I did some research on my own after learning about conditionally convergent sequences. I have become interested in learning about infinite products. From my research, it seems that a criteria of a convergent infinite product is that the sequence converges to 1. This sequence does converge to 1 so it meets this criteria. I believe I need to show that

$$\sum_{n=1}^{\infty} \ln\left(\frac{2n+3}{2n+2}\right)$$

converges. I tried to prove this using the ratio test, but I got nowhere with that. Any suggestions?

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    $\begingroup$ Do you know what the linear approximation is for $\ln$ at $1$? I.e., $\ln(1+x) \approx \ $? $\endgroup$ – Ted Shifrin Dec 5 '16 at 1:05
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    $\begingroup$ Possible duplicate of How to prove that $\lim_{n \to\infty} \frac{(2n-1)!!}{(2n)!!}=0$ $\endgroup$ – Simply Beautiful Art Dec 5 '16 at 1:07
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    $\begingroup$ @ArnaldoNascimento That doesn't work. $\endgroup$ – Simply Beautiful Art Dec 5 '16 at 1:08
  • $\begingroup$ Note that there aren't very many infinite product questions on this site, and so it only takes a few minutes to briefly check if your question is a duplicate. $\endgroup$ – Simply Beautiful Art Dec 5 '16 at 1:10
  • $\begingroup$ All the terms are greater than 1. Product of such infinite numbers greater than 1 has to be infinite. $\endgroup$ – jnyan Dec 5 '16 at 16:37
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The series diverges by the integral test. Consider $f(x) = \log\left(\frac{2x+3}{2x+2}\right) = \log (2x+3) - \log(2x+2)$ defined on $[1, +\infty)$.

Integrating this, by parts, yields the antiderivative$$ F(x)=\left(x+\frac32\right)(\log (2x+3)) - (x+1)\log(2x+2) = \frac 12(\log(2x+3)) +(x+1)(\log(2x+3)) - (x+1)(\log (2x+2)) = (x+1) \log\left(\frac{2x+3}{2x+2}\right) \ + $$

$$\frac 12\log(2x+3) \xrightarrow[x\to+\infty]{} +\infty$$

which implies $\int_1^{+\infty} f(t) \ dt = \lim_{x \to \infty}\left(F(x) - F(1)\right) = + \infty $.

Of course, this implies the product diverges as well.

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  • $\begingroup$ Thank you, this answered my question. $\endgroup$ – MathGuy Dec 5 '16 at 20:20
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \prod_{n = 1}^{\infty}\pars{2n + 3 \over 2n + 2} &= \lim_{N \to \infty}\,\,\prod_{n = 1}^{N}\pars{n + 3/2 \over n + 1} = \lim_{N \to \infty}\,\,{\pars{5/2}^{\overline{N}} \over \pars{2}^{\overline{N}}} = \lim_{N \to \infty}\,\,{\Gamma\pars{5/2 + N}/\Gamma\pars{5/2} \over \Gamma\pars{2 + N}/\Gamma\pars{2}} \end{align}

As $\ds{N \to \infty}$, the Stirling Asymptotic Formula yields: \begin{align} {\Gamma\pars{5/2 + N} \over \Gamma\pars{2 + N}} &\sim {\root{2\pi}\pars{5/2 + N}^{N + 3}\,\,\expo{-\pars{N + 5/2}} \over \root{2\pi}\pars{2 + N}^{N + 5/2}\,\,\expo{-\pars{N + 2}}} = \root{N}\,{\bracks{1 + \pars{5/2}/N}^{N }\,\,\expo{-1/2} \over \pars{1 + 2/N}^{N}} \\[5mm] & \sim \root{N}\,{\expo{5/2}\expo{-1/2} \over \expo{2}} = \root{N} \end{align}


$$ \mbox{Then,}\quad\bbx{\ds{% \prod_{n = 1}^{N}\pars{n + 3/2 \over n + 1} \sim {4 \over 3\root{\pi}}\,\root{N} \quad\mbox{as}\quad N \to \infty}} $$

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  • $\begingroup$ Seems a bit overkill just to determine convergence :). $\endgroup$ – Erick Wong Dec 5 '16 at 2:04
  • $\begingroup$ @ErickWong I agree with you. $Stirling$ is the first 'thing' which comes to our mind. Thanks. $\endgroup$ – Felix Marin Dec 5 '16 at 2:43
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A First Approach

Since $(2n+3)^2\gt(2n+3)^2-1=(2n+2)(2n+4)$, we get $$ \frac{2n+3}{2n+2}\ge\frac{2n+4}{2n+3} $$ Therefore, we have $$ \begin{align} \left(\prod_{n=1}^m\frac{2n+3}{2n+2}\right)^2 &\ge\prod_{n=1}^m\frac{2n+3}{2n+2}\frac{2n+4}{2n+3}\\ &=\prod_{n=1}^m\frac{2n+4}{2n+2}\\ &=\frac{2m+4}4 \end{align} $$ Thus, $$ \prod_{n=1}^m\frac{2n+3}{2n+2}\ge\sqrt{\frac{m+2}2} $$ and the infinite product diverges.


A Second Approach

For $a,b\ge0$, we have $(1+a)(1+b)\ge1+a+b$. Therefore, by induction, we get $$ \prod_{n=1}^m(1+a_n)\ge1+\sum_{n=1}^ma_n $$ Thus, $$ \begin{align} \prod_{n=1}^m\left(1+\frac1{2n+2}\right) &\ge1+\sum_{n=1}^m\frac1{2n+2}\\ &=1+\frac12\sum_{n=2}^{m+1}\frac1n \end{align} $$ which diverges by comparison with the Harmonic Series.

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Two useful general results about real sequences:

(1). If $a_n\geq 0$ for all $n$ then $\prod_{n=1}^{\infty}(1+a_n)<\infty \iff \sum_{n=1}^{\infty}a_n<\infty.$

(2). If $0\leq a_n<1$ for all $n$ then $\prod_{n=1}^{\infty}(1-a_n)\ne 0\iff \sum_{n=1}^{\infty}a_n<\infty.$

These are not hard to prove. A useful part of proving (2) is that if the product is not $0$ then $0\leq a_n<1/2$ for all but finitely $n.$

Apply (1) with $a_n=1/(2n+2).\;$We have $\sum_{n=1}^{\infty}a_n=\infty.$ Therefore $\prod_{n=1}^{\infty}(2n+3)/(2n+2)=\prod_{n=1}^{\infty}(1+a_n)=\infty.$

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Just as a complement to the answer by Felix (being too long for a comment thereto), consider that we may resort to the definition of the Partial Gamma, as: $$ \begin{gathered} \Gamma (z) = \mathop {\lim }\limits_{n\, \to \,\infty } \left( {\Gamma _{\,n} (z) = n^{\,z} \frac{{\,n!}} {{z^{\,\overline {\,n + 1\,} } }} = n^{\,z} \frac{1} {z}\prod\limits_{1\, \leqslant \,k\, \leqslant \,n} {\frac{k} {{z + k}}} } \right) \hfill \\ \mathop {\lim }\limits_{n\, \to \,\infty } \prod\limits_{1\, \leqslant \,k\, \leqslant \,n} {\frac{k} {{z + k}}} = \mathop {\lim }\limits_{n\, \to \,\infty } \,\frac{{z\,\Gamma (z)}} {{n^{\,z} }} \hfill \\ \end{gathered} $$ wherefrom we get $$ \begin{gathered} \prod\limits_{1\, \leqslant \,n} {\frac{{2n + 3}} {{2n + 2}}} = \mathop {\lim }\limits_{N\, \to \,\infty } \prod\limits_{1\, \leqslant \,n\, \leqslant \,N} {\frac{{\left( {n + 3/2} \right)}} {{\left( {n + 1} \right)}}} = \mathop {\lim }\limits_{N\, \to \,\infty } \frac{{\prod\limits_{1\, \leqslant \,n\, \leqslant \,N} {\frac{n} {{\left( {n + 1} \right)}}} }} {{\prod\limits_{1\, \leqslant \,n\, \leqslant \,N} {\frac{n} {{\left( {n + 3/2} \right)}}} }} = \hfill \\ = \mathop {\lim }\limits_{N\, \to \,\infty } \frac{{1\,\Gamma (1)}} {{N^{\,1} }}\tfrac{{2\,N^{\,3/2} }} {{3\,\Gamma (3/2)}} = \frac{2} {{3\,\Gamma (3/2)}}\mathop {\lim }\limits_{N\, \to \,\infty } \sqrt N \hfill \\ \end{gathered} $$

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In fact if $a_n>0,$ then $\prod_{n=1}^\infty (1+a_n)$ converses iff $\sum a_n$ converges. You can obtain this result by applying the logarithm and using $\ln (1+u) \sim u$ as $u\to 0.$

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  • $\begingroup$ And this was downvoted why? $\endgroup$ – zhw. Dec 5 '16 at 18:46

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