Old title:

Is it always true that a sum to some $m$ equals the limit of that sum to some $x$ such that $x$ tends to $m$?
$\ $

This may seem a stupid question, but I'm placing it because although it seems obviously true, it may not be so in some edge cases. After all, all true statements in mathematics are obvious (or at least analytic and/or tautological propositions).

Also, sorry for the long question title, but I didn't know how to phrase it without using "this".

$$ \forall f,\ \forall n\in\mathbb{N}\cup \{0\},\ \forall m\in\mathbb{N}\cup\{0; \ \infty\},\ \forall x \in D_f, \sum_n^mf(x)=\lim_{x\to m}\sum_{n}^{x}f(x) $$

Notice that often, $x = n$, they can be the very same variable. So, is the above true (under any circumstances, except wrongness, of course)?

Added: Note: I prefer not to state much about $x$ because even being an upper limit of a sum, it may be non-integer by analytic extension (I don't know if this is the correct name, but I think you get the idea) of the operation, in the limit, as $x$ approaches $m$, it may be a fractional number, perhaps requiring you to (if such is the function) do a half and then to do two thirds of a derivative, and add that to the previous complete derivatives, studying how the partial sums approach the final value.


In the following examples the stated above applies.

Where $x=n$

$$ \sum_{n=0}^{\infty}\left(9\times 10^{-n}\right)=\lim_{x\to \infty}\sum_{n=0}^{x}\left(9\times 10^{-n}\right)=10 $$

Where $x\not=n$

$$ \sum_{n=0}^5 3=\lim_{x\to 5}\sum_{n=0}^{x}3=5\times 3=15 $$

Thank you in advance.

Very Late Relevant Edit

I have found doubtful cases and reminded of this question, such as the one shown here.

It seems quite obvious from evaluating partial sums that

$$ \lim_{n\to\infty}\sum_{i=0}^n2^n=\infty $$


$$ \sum_{i=0}^\infty2^n=2^0+\sum_{i=1}^\infty2^n=1+2\sum_{i=0}^\infty2^n \iff\\ \iff-\sum_{i=0}^\infty2^n=1\iff\sum_{i=0}^\infty2^n=-1\neq\lim_{n\to\infty}\sum_{i=0}^n2^n $$

Which seems to answer "no" to my question. Unless I'm missing something. Am I? Is this logic correct?

  • 1
    $\begingroup$ What do you mean by $\sum_n^x$ when $x$ is not an integer? And if $D_f$ only consists of integers, it's still not clear what you mean by the limit on the right hand side. $\endgroup$ – mrf Jun 11 '13 at 18:53
  • $\begingroup$ @mrf, thanks for the warning, I hadn't noticed it yet, see the updated post. $\endgroup$ – JMCF125 Jun 11 '13 at 19:38
  • $\begingroup$ The definition seems fine, but (in very late edit) the manipulation using the symbol $\infty$ does not. $\endgroup$ – André Nicolas Mar 14 '14 at 20:28
  • $\begingroup$ @AndréNicolas, what do you mean by manipulation of $\infty$? I didn't do any "tricks" with it. And by definition, you mean the definition of what exactly? Is what I added in the edit invalid? $\endgroup$ – JMCF125 Mar 15 '14 at 12:27
  • $\begingroup$ On the next to last displayed line, the "taking out" a common factor of $2$ is slightly problematic, and the subtraction that leads to the beginning of the next line is definitely problematic. $\endgroup$ – André Nicolas Mar 15 '14 at 16:21

If you mean $\sum_n^m f(x) = \sum_{x=n}^m f(x)$, then the formula you wrote is true. It is true for $m = \infty$ just by the definition of the infinite series, and for finite $m$ due to the fact that $\lim_{k}x_k = m$ for the integer sequence of $x_k$ if and only if $x_k = m$ for all $k$ big enough.

  • $\begingroup$ What if it is a sum of fixed values ($x\not = n$, thus my definition to cover this case)? $\endgroup$ – JMCF125 Jun 10 '13 at 15:44
  • $\begingroup$ @JMCF125: can you provide an example, please? $\endgroup$ – Ilya Jun 10 '13 at 15:46
  • $\begingroup$ Any case where $x$ is fixed and $n$ is variable. $\endgroup$ – JMCF125 Jun 10 '13 at 16:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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