Additional insights when converting sums to products Given some sum, $$\displaystyle\sum \ln x_i =  k $$ We have $$\ln \prod x_i = k$$
I've always found this relation to be really interesting.  I saw it used once in a linear algebra proof but I haven't seen it since.  Are there any other interesting uses of this trick?
 A: Yes, finding product expansion for sums, especially infinite sums, can sometimes shed new light on them. For example, consider the zeta function, defined for complex numbers with $\Re(z)>1$ by 
$$\zeta(s)=\sum_{n=0}^\infty \frac{1}{n^s}.$$
It is not hard, though not trivial, to show this "factors" as one would expect, giving the infinite product
$$\zeta(s)=\prod_p^\infty \left(1-\frac{1}{p^s}\right)$$
where the product is taken over all primes $p$.
This shows that the zeta function has no zeros for $\Re(z)>1$, something which is not obvious from examining the sum representation.
A: I often use the series of the logarithm to reduce computational effort when computing logarithms.  It is usually easier to compute many small logarithms then one large one, so I often exploit this by cutting up the number I want to compute.
This is a simple way to compute $\log x$.  We have
$$\log x=\log \left(\frac {x}{2}\right)+\log 2$$
repeating this process $n$ times with a pre-computed $\log 2$, we have
$$\log x=\log \left(\frac{x}{2^{n}}\right)+n\log 2$$
We can then quickly compute $\log \left(\frac{x}{2^{n}}\right)$ with a Taylor's Series, its Chebyshev economized counterpart, or a different, more efficient method.  I usually choose $n$ as a number that makes $\left(\frac{x}{2^{n}}\right) \approx 1$ to make it easier to estimate in a series.

For example, I may compute $\log 10$ efficiently using the above basic method.  Say we know $\log 2$ to $10$ places, i.e. $0.6931471805$.  Letting $n=3$ (making $\left(\frac{10}{2^{n}}\right) = 1.25$), 
$$\log 10=\log \left( \frac{10}{8} \right)+3 * 0.6931471805=\log \left( \frac{5}{4} \right)+2.0794415415$$
Already, note that $2.0794415415$ is a decent estimate of $\log 10$. Using 5 terms of the basic Taylor's series expnsion of $\log$, we have
$$\log \left( \frac{5}{4} \right)\approx \left( \frac{5}{4} -1\right)-\frac{1}{2}\left( \frac{5}{4} -1\right)^2+\frac{1}{3}\left( \frac{5}{4} -1\right)^3-\frac{1}{4}\left( \frac{5}{4} -1\right)^4+\frac{1}{5}\left( \frac{5}{4} -1\right)^5 \approx 0.22317708\bar{3}$$
accurate to 4 decimal places.  Together, we have
$$\log 10\approx 0.223177083+2.0794415415=2.3026186245$$
accurate again to 4 decimal places.
