Prove that the infinite product (2n+3)/(2n+2) converges or diverges. 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?
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\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}

$$
\mbox{Then,}\quad\bbx{\ds{%
\prod_{n = 1}^{N}\pars{n + 3/2 \over n + 1} \sim {4 \over 3\root{\pi}}\,\root{N}
\quad\mbox{as}\quad N \to \infty}}
$$
A: 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.
A: 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. 
A: 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 $\prod_{n=1}^{\infty}(2n+3)/(2n+2)=\prod_{n=1}^{\infty}(1+a_n)=\infty.$
A: 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
$$
\begin{gathered}
  \prod\limits_{1\, \leqslant \,n} {\frac{{2n + 3}}
{{2n + 2}}}  = \mathop {\lim }\limits_{N\, \to \,\infty } \prod\limits_{1\, \leqslant \,n\, \leqslant \,N} {\frac{{\left( {n + 3/2} \right)}}
{{\left( {n + 1} \right)}}}  = \mathop {\lim }\limits_{N\, \to \,\infty } \frac{{\prod\limits_{1\, \leqslant \,n\, \leqslant \,N} {\frac{n}
{{\left( {n + 1} \right)}}} }}
{{\prod\limits_{1\, \leqslant \,n\, \leqslant \,N} {\frac{n}
{{\left( {n + 3/2} \right)}}} }} =  \hfill \\
   = \mathop {\lim }\limits_{N\, \to \,\infty } \frac{{1\,\Gamma (1)}}
{{N^{\,1} }}\tfrac{{2\,N^{\,3/2} }}
{{3\,\Gamma (3/2)}} = \frac{2}
{{3\,\Gamma (3/2)}}\mathop {\lim }\limits_{N\, \to \,\infty } \sqrt N  \hfill \\ 
\end{gathered} 
$$
A: 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 $u\to 0.$
