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Re: Re: Sum


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For example:

i = 3;
Sum[1/i^2, {i, 1, Infinity}]

Pi^2/6

while

Sum[Evaluate[1/i^2], {i, 1, Infinity}]

Infinity

Andrzej Kozlowski

On 17 Mar 2005, at 09:29, DrBob 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
>
>


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