integral test to show that infinite product $ \prod \limits_{n=1}^\infty\left(1+\frac{2}{n}\right)$ diverges 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.

If You dont have to use integration test then You can do something like this, it is much easier.
$$\prod\limits_{n=1}^{k}(1+\frac{2}{n})=\frac{(k+1)(k+2)}{2}.$$
It is very easy to show since it is equal to: $\frac{1}{3}\cdot\frac{2}{4}\cdot\frac{3}{5}\cdot\frac{4}{6}\cdot\frac{5}{7}\cdot\cdot\cdot\cdot$
You can reduce almost all this fractions. Denominator of nth fraction can be reduced with numerator of n+2th fraction.
