logarithmssummation
In working through a problem, I've encountered the need to express
$\log n = \sum \limits_{k = 1}^{n – 1} \log(1 + \frac{1}{k})$
where $\log k $ is the natural logarithm of k. I'm fairly certain the above equality holds (please let me know if otherwise), but I'm not quite sure how to derive it. Any help would be greatly appreciated.
So far, I've tried expressing $\log n = \log(n!) – \log((n-1)!)$ to no avail.
Many thanks!
One way to approach this question is to consider the minimum of $x - \log_a x$ on the interval $(0,\infty)$. For this we can compute the derivative, which is $1 - 1/(\log_e a )\cdot x$. Thus the derivative is zero at a single point, namely $x = 1/\log_e a,$ and is negative to the left of that point and positive to the right. Thus $x - \log_a x$ decreases as $x$ approaches $1/\log_e a$ from the left, and then increases as we move away from this point to the right. Thus the minimum value is achieved at $x = 1/\log_e a$. (Here I'm assuming that $a > 1$, so that $\log_e a > 0$; the analysis of the problem is a little different if $a < 1$, since then for $x < a < 1$, we have $log_a x > 1 > x,$ and the statement is not true.)
Now this value is equal to $1/\log_e a + (\log_e \log_e a)/\log_e a,$ and you want this to be $> 0 $. This will be true provided $a > e^{1/e}$ (as noted in the comments).
The way we define the Logarithm is the inverse of the exponential.
If $z= re^{i\theta}$,then $ln(z)=ln(r)+i\theta$.
Now , a problem that is there is, with this equation alone , logarithm is multi-valued i.e for every z there are infinitely many values for the logarithm . For $\theta$ differing by a value of $2\pi$ , one will get the same value. It isn't injective. This leads us restricting its domain so that each point(z) corresponds to only one value. This is referred to as a branch cut. Check the image from Wiki,
So, for log(z), simply remove any ray joining 0 and infinity.(Restricting complete rotation(0 to $2\pi$!). Also, see in the graph what happens. The principle log removes one particular ray, the negative real axis, I guess. This is only a convention, I guess.
Now, coming back to your question, you want log(iz)(PS:-The $iz$ only changes the labels on the axes) to be analytic in some given region. All you need to do is to remove a ray(not in the given region) to make it analytic in the given region.
Best Answer
It telescopes after you do a little algebra:
$$\begin{align*} \sum_{k=1}^{n-1}\log\left(1+\frac1k\right)&=\sum_{k=1}^{n-1}\log\frac{k+1}k\\\\ &=\sum_{k=1}^{n-1}\big(\log(k+1)-\log k\big)\\ &=\log n-\log 1\\ &=\log n\;. \end{align*}$$