what is the summation of \( \sum_{n=1}^\infty \log\left(\dfrac{n(n+2)}{(n+1)^2}\right)\)and \(\displaystyle\sum_{n=1}^{\infty} (\sqrt[n]{2} - 1)\)?
+505 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
