What is the meaning of adding/subtracting a fraction in a summation limit inside a limit?

Question

In Wolfram Alpha entry 1 the following is claimed to be true:

$$\lim_{n\to \infty } \left(\sum _{k=1}^n \log (2 k)-\sum _{k=1}^{n+\frac{1}{2}} \log (2 k-1)\right)=\frac{1}{2} \log \left(\frac{\pi }{2}\right)$$

What is the meaning of adding a fraction like $\frac{1}{2}$ in the upper summation limit in the second sum, inside the limit?

Is it a bug a or a feature?

Compare to Wolfram Alpha entry 2 where the fraction $\frac{1}{2}$ in the upper summation limit of the second sum has been left out:

$$\lim_{n\to \infty } \left(\sum _{k=1}^n \log (2 k)-\sum _{k=1}^{n} \log (2 k-1)\right)=\infty$$

I get the same results in Mathematica 8.0.1:

Limit[Sum[Log[2*k], {k, 1, n}] - Sum[Log[2*k - 1], {k, 1, n + 1/2}], 
 n -> Infinity]
Limit[Sum[Log[2*k], {k, 1, n}] - Sum[Log[2*k - 1], {k, 1, n}], 
 n -> Infinity]

and as well as:

Limit[Sum[Log[2*k], {k, 1, n - 1/4}] - Sum[Log[2*k - 1], {k, 1, n + 1/4}], n -> Infinity]
Limit[Sum[Log[2*k], {k, 1, n}] - Sum[Log[2*k - 1], {k, 1, n + 1/4}], n -> Infinity]

The only support I was able to find in the documentation for Sum is:

"The limits of summation need not be numbers. They can be Infinity or symbolic expressions."

Answer 1

Evaluate the sums without Limit, so that you understand what limit is being computed. The problem has nothing to do with Limit. It's a bug in Sum[].

Sum[Log[2*k - 1], {k, 1, a}]
% /. a -> n + 1/2
% == Sum[Log[2*k - 1], {k, 1, n + 1/2}]
(*
Log[2^a Pochhammer[1/2, a]]
Log[2^(1/2 + n) Pochhammer[1/2, 1/2 + n]]
True
*)

Arguably, it's a "bug" in the user input. A full specification at least has non-erroneous behavior (does not evaluate):

Sum[Log[2*k - 1], {k, 1, n + 1/2}, 
 Assumptions -> n \[Element] Integers]
(*
Sum[Log[-1 + 2 k], {k, 1, 1/2 + n}, 
 Assumptions -> n \[Element] Integers]
*)

Answer 2

Limit is assuming that n is continuous, and so the sums are simplified to give results for non-integers. You can see this with TracePrint, it eventually evaluates:

Log[2^n Pochhammer[1,n]]-Log[2^(1/2+n) Pochhammer[1/2,1/2+n]]

This function does converge. However, it doesn't equal the sums for any n (the first Log[...Pochhammer... agrees with the first sum on $\mathbb Z$, and the second agrees with the second sum on $\mathbb Z+\frac12$, but together they don't agree anywhere).

I don't know how to make the Limit evaluate if you explicitly write Floor[n+1/2] in the bound, but Roland's comment is correct.

The story is the same for your other functions (although the last pair diverges even after 'continuousization').