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# calc questions im lost on

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what is the summation of $$\sum_{n=1}^\infty \log\left(\dfrac{n(n+2)}{(n+1)^2}\right)$$and $$\displaystyle\sum_{n=1}^{\infty} (\sqrt[n]{2} - 1)$$?

Mar 31, 2021

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First one:

The rule of addition of logarithms is: $$\log(x)+\log(y)=log(xy)$$. Therefore, we need to calculate $$\displaystyle \prod_{n=1}^{\infty}\frac{n\left(n+2\right)}{\left(n+1\right)^{2}}$$ first, and then take the log of that infinite product.

The infinite product would look something like this:

$$\frac{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot4\cdot6 \dotsm}{2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot5\cdot5\dotsm}$$, and you can rearrange it like this:

$$\frac{1\cdot2\cdot3\cdot3\cdot4\cdot4\cdot5\cdot5\dotsm}{2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot5\cdot5\dotsm} = \frac{1}{2}$$ (since the numerator and denominator both contain $$2\cdot3\cdot3\cdot4\cdot4\cdot5\cdot5\dotsm$$)

Therefore, the answer to the original problem is $$\boxed{\log{\frac{1}{2}}}$$, which is approximately -0.301

Mar 31, 2021