Taylor series of natural log?
I have a weird sum and I figure that it should be Taylor series of natural log. However, I try my best but I can not transform it base on equation on wikipedia. Please help.
Need-to-transform sum: lim (n->infi) (sigma k: 1->n) [of] (1/(n+k))
(like 1/n + 1/(n+1) + 1/(n+2) + ... + 1/2n)
Suspected to be equal to: ln(1/2)
Taylor series of ln: ln (x) = (sigma n: 1->infi) [of] (-1)^(n+1) * (x-1)^n / n
Comments
The ln(x) function requires care because if you try to evaluate it at x = 0, those derivatives ( ~ x^{-n} for n > 0) will diverge. That is why the taylor series given looks the way it does. In fact, the way you get that Taylor series is you start with something else, and then modify it in the end so you get ln(x). This is what I mean,
a *very* well known series is 1 / (1 - x) = Sum x^k, from k = 0 to infinity, valid for |x| < 1, this is the radius of convergence. If this is not known to you, take a second to derive it from Taylor's theorem.
Now let x -> -x = (-1)x, the above becomes
1 / (1 + x) = Sum (-1)^k x^k
integrate both sides
integral dx / (1 + x) = integral sum (-1)^k x^k dx
An argument can be made that permits interchanging the order of integration and summation on the right-hand side (we can talk about it if you like, but I will just suggest at the moment that we accept it is true),
integral dx / (1 + x) = sum integral (-1)^k x^k dx
ln (1 + x) = Sum (-1)^k x^{k + 1} / (k + 1) for k = 0 to infinity
now let x --> x - 1, then ln(1 + x) = ln (1 + x - 1) = ln (x)
ln (x) = Sum (-1)^k (x - 1)^{k + 1} / (k + 1) for k = 0 to infinity
reindex, let n = k + 1 --> at k = 0 we have n = 1, and at k = infinity we have n = infinity, further note that this implies n = k + 1 --> k = n - 1, so (-1)^k = (-1)^(n - 1)
thus
ln (x) = Sum (-1)^{n - 1} (x - 1)^n / n for n = 1 to infinity
now, they have (-1)^{n + 1} and I have (-1)^{n - 1}, you should convince yourself that these are the same for any n, i.e. it takes on a value of only 1 (if the exponent is even or zero) and -1 (if the exponent is odd), so we can write it the way they have on wiki without consequence,
ln (x) = Sum (-1)^{n + 1} (x - 2)^n / n for n = 1 to infinity
ok, with that out of the way: is what you have equal to ln (1/2)?
you have Sum (1 / (n + k)) for k = 1 to n
first of all, this is a finite sum, so it can never be exactly equal to ln(1/2) as per out previous analysis.
We can stop there if that is satisfying, you need an infinite number of terms to get ln(1/2) and you do not, so you cannot show one equals the other.
If you want more convincing, try to seek ln(1/2) as an expansion,
recall ln (1 + x) = Sum (-1)^k x^{k + 1} / (k + 1) for k = 0 to infinity from our previous work,
so for |x| < 1 this is valid, we can make x = -1/2 = (-1) 1/2
ln(1 - 1/2) = ln (1/2) = Sum (-1)^k (-1)^{k+1} (1/2)^{k + 1} / {k + 1}
ln (1/2) = Sum (-1)^{2k + 1} (1/2)^{k + 1} / {k + 1}
now, 2k + 1 is *always* odd for any k (convince yourself), so (-1)^{2k + 1} = -1, this makes sense since ln(1/2) is negative,
ln (1/2) = - Sum (1/2)^{k + 1} / {k + 1}
reindex n = k + 1 -->
ln (1/2) = - sum (1/2)^n / n for n = 1 to infinity
Notice this does not look like what you have at all.
The word 'natural' is probably a misnomer. Base 10 is certainly easier to use for common calculations, but e comes into mathematics very soon and for many purposes e is the base that fits the mathematicians. It is all to do with the exponential function and all its ramifications. Consult a text book on this and a bit of math history. Napier did not use e as his base when he invented logarithms. Neither did he use base 10.