Re: Re: Re: Sum

```Yes, I see that now. Thanks!

Bobby

On Fri, 18 Mar 2005 05:33:52 -0500 (EST), yehuda ben-shimol <bsyehuda at gmail.com> wrote:

> As said before, the iterator of Sum is a LOCAL variable to the Sum
> function as can be seen in the following example
> Sum[(Print[i]; 1/i^2), {i, Infinity}]
> and the functions prints
> K\$18
> and then returns the result
> Pi^2/6
> K\$18 is a local naming (as is used to in a Module[] for local
> variables) so x must be evaluated with the local variables before the
> summation takes place
>
> yehuda
>
>
> On Thu, 17 Mar 2005 03:29:08 -0500 (EST), DrBob <drbob at bigfoot.com> wrote:
>> >> If the result weren't infinite, that would be a good idea.
>>
>> Oops, I take that back. It would almost always be a bad idea, so I don't know why Evaluate isn't automatic.
>>
>> Bobby
>>
>> On Wed, 16 Mar 2005 10:10:40 -0600, DrBob <drbob at bigfoot.com> wrote:
>>
>> > Sum tried to simplify BEFORE substituting the value of x. If the result weren't infinite, that would be a good idea. But in this case, it's not. But there's a simple fix:
>> >
>> > x = Exp[-k]
>> > Sum[Evaluate@x, {k, 0, Infinity}]
>> >
>> > E^(-k)
>> > E/(-1 + E)
>> >
>> > Bobby
>> >
>> > On Wed, 16 Mar 2005 05:35:51 -0500 (EST), mjumbo <mjumbo at nm.ru> wrote:
>> >
>> >> Can sombody explain me why
>> >>
>> >> Sum[Exp[- k], {k, 0, Infinity}]
>> >>
>> >> gives the correct result, while the following (imho equivalent) statements
>> >>
>> >> x = Exp[-k]
>> >> Sum[x, {k, 0, Infinity}]
>> >>
>> >> don't? What should I do to make it work? It is rather vexing to copy/paste long expressions instead of using short placeholders like x.
>> >> __________
>> >> www.newmail.ru -- Íîâàÿ Ïî÷òà: âñå ïî-íîâîìó.
>> >>
>> >>
>> >>
>> >>
>> >
>> >
>> >
>>
>> --
>> DrBob at bigfoot.com
>>
>>
>
>
>
>

--
DrBob at bigfoot.com

```

• References:
• Sum
• From: mjumbo <mjumbo@nm.ru>
• Re: Re: Sum
• From: yehuda ben-shimol <bsyehuda@gmail.com>
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