1
$\begingroup$

I've had this confusion for quite sometime now. Let's say a term/summand in a series is undefined / tending to infinity. Further, It is possible to prove convergence of the series and also possible to evaluate the series. Now how do we decide if we should conclude that series is undefined or if the series is evaluatable.I think this can be thought of as a sister-confusion to whether $\int_0 ^a \frac{dx}{2\sqrt x}$ is taken to be undefined or $\sqrt a$ except I'm clear with that.

Here is one such problem for discussion (I'm certain the description above is insufficient to really convey what the subject of this question is and thus we can possibly use this problem):

$$f(x) = \sum _{k=1}^{n} \frac {x}{(kx+1)(kx + 1 - x)} $$

I have two orders of evaluation:

1. Evaluating Summation First:

This is easily telescopable and gives $1 - \frac {1}{nx+1}$

Now this gives than only discontinuity is $x = \frac {-1}n$

What's happening here is when a term is undefined, another term is also undefined and their sum isn't . Like at $x=-1$, $$\frac {x}{(x+1)(1)} + \frac x{(2x+1)(x+1)} = \frac {(x)(2x+2)}{(1)(x+1)(2x+1)} = \frac {2x}{2x+1}$$

2. Evaluating the Terms first:

If evaluating Summands/Terms first, we have at $x=-1$ $$\text {undefined} + \text {undefined} + \frac 12 + \frac 16...$$

Now this is undefined.

So we get discontinuities at many more points.

Question: is how to decide which way/order of evaluation to go for?

I can see that this is similar to the difference between $$\lim _{x \to 0} \frac {\sin x-x}{x^3}$$ and $$\lim _{x \to 0} \frac {\sin x}{x^3} - \lim _{x \to 0} \frac {x}{x^3}$$ but in this case the limits clearly tell us the order of evaluation, which I'm not able to see in the series of discussion (or any other series with a similar scenario)

Thanks

$\endgroup$
1
  • $\begingroup$ You can’t put $x=-1$ in $f(x)$ as $x=-1$ is not in domain of $f$ so $f$ is not defined at $x=-1$. $\endgroup$
    – Koro
    Commented Apr 14, 2021 at 15:17

0

You must log in to answer this question.