Take the 2-minute tour ×
Mathematica Stack Exchange is a question and answer site for users of Mathematica. It's 100% free, no registration required.

When I declare the function $f$ as follows:

f[n_] := Sum[x/Sum[Sqrt[y], {y, 1, x}], {x, 1, n}]/n

the input

Limit[f[n],n -> Infinity]

does not yield an answer, in fact the evaluation is terminated after a few seconds. If I now write

Limit[f[x], x -> Infinity]

I get an answer of $0$. Why is that, according to the manual entry of "Sum" we have

The iteration variable i is treated as local, effectively using Block.

If it matters, I use the version 9.0.

share|improve this question
add comment

1 Answer

up vote 3 down vote accepted

Beware of held attributes. Sum uses Block but also has HoldAll, which means that nothing is evaluated until the variable gets a value inside it - if any at all. Limit does not have held attributes.

If after defining f you evaluate f[n] (assuming n doesn't have a value in your environment) you get a

$$\frac{\sum _{x=1}^n \frac{x}{H_x^{\left(-\frac{1}{2}\right)}}}{n}$$

whereas f[x] (again, assuming x is free) gives you

$$\frac{\sum _{x=1}^x \frac{x}{H_x^{\left(-\frac{1}{2}\right)}}}{x}$$

So, note now the different summation limits and how what used to be a free variable is now colliding with the bounds. That shows what Limit actually gets and the answer, I hope.

share|improve this answer
    
I understand, it still feels weird that they implemented it that way. In my opinion taking a limit of some function should not depend on how you name the variable which is considered for the limit. –  Listing Jan 6 at 17:08
1  
When it comes to manipulating expressions, one needs to be aware of whether a built-in has or not hold behaviour as that affects how it works. And some expression handling built-ins don't have that (notably Integrate and D), allowing you to generate a expression before handing that over to the built-in, but others do. That's the way it is. –  caya Jan 6 at 17:14
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.