This is part of an assignment that I need to get a good mark for - I'd appreciate it if you guys could look over it and give some pointers where I've gone wrong.
(apologies for the italics)
$\prod_{n=1}^\infty\left(1+\frac{2}{n}\right)\; \text{ converges when } \sum_1^\infty \ln\left(1+\frac{2}{n}\right)\; \text{ converges }.$ $\sum_1^\infty \ln\left(1+\frac{2}{n}\right)\;=\;\sum_1^\infty \ln(n+2)-\ln(n)$ $ \text{let } f(x)=\ln(x+2)-\ln(x) \rightarrow f'(x)=\frac{1}{x+2} - \frac{1}{x}$ $ = \frac{x-x-2}{x(x+2)} = \frac{-2}{x(x+2)}<0$
$f(x)\ \text{is a decreasing function}.$ $f(x) \; \text{is a positive function for} \;x\geq1$ $f(x)\;\text{is a continuous function for} \;x>=1$
using integration test. $\int_1^\infty \ln(x+2) - \ln(x) = \lim_{t \to \infty}\int_1^t \ln(x+2)dx - \lim_{t \to \infty}\int_1^t \ln x dx$
$\int \ln(x)dx = x \ln x - x + c \Rightarrow \int \ln(x+2) = (x+2)\ln(x+2) - (x+2) + c$ Therefore $\int \ln(x+2) - \ln(x)dx = (x+2)\ln(x+2)-x - 2 - x \ln(x) + x + c$ $ = x \ln(\frac{x+2}{x})+ 2\ln(x+2)-2 + c $ Therefore, $\int_1^\infty \ln(x+2) - \ln(x)dx = \lim_{t \to \infty}\left[x \ln(\frac{x+2}{x}) + 2 \ln(x+2) - 2\right]_1^t$
$ = \lim_{t \to \infty}\left[t \ln(\frac{t+2}{t}) + 2\ln(t+2) - 2\right] - \lim_{t \to \infty}\left[\ln(\frac{3}{1}) + 2\ln(3) - 2\right] $
$ =\lim_{t \to \infty}\left[t \ln(\frac{t+2}{t}) + 2\ln(t+2) - 3\ln(3)\right]$
$ As\; t\rightarrow\infty, \; \lim_{t \to \infty}t \ln\left(\frac{t+2}{t}\right) + 2\ln(t+2) = \infty. $
Therefore the series $\sum_1^\infty \ln\left(1+\frac{2}{n}\right) $ is divergent.
Similarly the infinite product $\prod_{n=1}^\infty\left(1+\frac{2}{n}\right)$ is also divergent.