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I hypothesize that the following category of divergent summations cannot be assigned finite values.

$$\lim_{w\to\infty}\sum_{n=0}^wa_n=\pm\infty$$

$$\lim_{n\to\infty}a_n-a_{n+1}=0$$

If the summation diverges to $\pm\infty$ and increases by less and less from $a_n$ to $a_{n+1}$, then we cannot assign a finite value to it.

I can't tell you why I think this is true other than my observations on common divergent series and the finite values we assign to them.

This is in contrast to the following summations, which can be assigned finite values and do not fall under my category:

$$\sum_{n=0}^{\infty}(-1)^n=1/2$$

$$\sum_{n=1}^{\infty}(-1)^{n+1}n=1/4$$

If you could provide a counter example for when we can assign a finite value for divergent summations of my type, please do so.

Here is my intuitive explanation for why my summations happen this way.

Imagine as though, for a moment, that $+\infty$ and $-\infty$ were one and the same thing. This allows us to take the following limit without worry:

$$\lim_{x\to0}\frac1x=\infty$$

Instead of a positive or negative infinity, it is one, whole infinity.

Now, as for the application in divergent summations that have partial sums always increasing.

As you take the following summation, you will find that it gets larger and larger and that the amount that it increases gets larger and larger:

$$\sum_{n=1}^{\infty}n=1+2+3+\dots$$

In fact, it gets so large, one of its large partial sums will reach infinity:

$$\lim_{i\to\infty}\sum_{n=1}^in\to\infty$$

But, intuitively, it will keep increasing, past a point of infinity, and become negative. In a way, this is very similar to the graph of $\frac1x$.

As we go from the right (positive) side of the graph of $\frac1x$, we see that it gets increasingly large. When it gets to $x=0$, the graph is infinite in both the up and down direction. As we continue to go left along the graph, we see that it starts at "$-\infty$", which is the same as $+\infty$, and it goes up from $-\infty$, much the same way it was originally moving. The entire time, it was going up, to infinity, then to the negatives. After a while, the graph stops increasing as we go left and converges to a finite value.

In relation to some divergent summations, you can abstractly think that they went to $+\infty$ and then exceeded it, coming around as a negative answer, which explains how it can be negative in the first place.

However, a summation such as $\sum_{n=1}^{\infty}\frac1x$, stops increasing after reaching infinity. And such is my intuitive reasoning to why such a thing should occur.

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  • $\begingroup$ I only note this through examples I have found. So counter-examples would do nicely. $\endgroup$ – Simply Beautiful Art Jan 26 '16 at 23:24
  • $\begingroup$ with assigning a finite value to a divergent series you mean $\lim_{x \to 1^-} \sum_n a_n x^n$ ?? en.wikipedia.org/wiki/Abel%27s_theorem#Remarks $\endgroup$ – reuns Jan 26 '16 at 23:33
  • $\begingroup$ @user1952009 I mean I have assigned finite values to divergent series using summation methods like Riemann sums and the zeta function and all that. You can get a feel for it in the divergent series (not the book) Wikipedia. $\endgroup$ – Simply Beautiful Art Jan 26 '16 at 23:35
  • $\begingroup$ so you mean the whole analytic continuation summation methods $\endgroup$ – reuns Jan 26 '16 at 23:38
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    $\begingroup$ with the en.wikipedia.org/wiki/Divergent_series#Abelian_means $\lim_{s \to 0^+} f(s) = \sum a_n n^{-s}$ the series $\sum_n n^b$ becomes summable for any $b \ne 0$ (instead of $b < -1$), so a counter-example to your condition is $\sum_n \frac{1}{\sqrt{n}}$ whose value is $\zeta(1/2)$ with the Abelian means summation method. $\endgroup$ – reuns Jan 27 '16 at 0:02

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