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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.

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    I'm very confused. It seems to me like the telescoping test shows convergence. If it converges, it converges absolutely, thus we can rearrange the terms. Then $\sum_1^\infty \ln(n+2)-\ln(n) = \ln(3)-\ln(1)+\ln(4)-\ln(2)+\ln(5)-\ln(3)+\dots$ which evaluates to $-\ln(1)-\ln(2) = -\ln(2)$. What am I doing wrong?2012-08-01

2 Answers 2

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For $a_n \ge 0 $ the infinite product $\prod_{n=1}^\infty (1+ a_n)$ converges precisely when the infinite sum $\sum_{n=1}^\infty a_n $ converges, since

$1+ \sum_{n=1}^N a_n \le \prod_{n=1}^N (1+ a_n) \le \exp \left( \sum_{n=1}^N a_n \right) . $

So you only need consider $ \sum_{n=1}^\infty 2/n $ and you can use your integral test for that, or just quote the standard result.

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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.