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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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    You can put text that you don't want in math format in \mtext (or, even better, just take it out of math mode).2010-12-23
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    Silly question: Is the answer wrong (that the product diverges), or is the proof wrong? Or neither?2010-12-23
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    Because that series really should diverge by telescoping...2010-12-23
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    I think he wants us to show the divergence by applying the integral test.2010-12-23
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    As Jesse said, the most sensible way to show that the series diverges is by telescoping. A meta remark: the fact that you need to get a good mark on the assignment is pretty irrelevant, so please leave it out next time or put it at the end if you can't resist.2010-12-23
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    Hi Alex, thanks for the response. sorry if the comment offended. I've used this method because the exact wording of the question states; Use the integral test to show that the infinite product diverges (hint - think about the properties of the logarithmic function).2010-12-23
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    The comment didn't offend, it is just irrelevant and obstructs the more relevant parts. E.g. the excerpt of the question on the main site reveals "Hi Guys, 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". Instead, it should be telling the users what the question is about (luckily, in this particular instance, your title already does a good job).2010-12-23
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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.