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I've tried to solve a problem and after some research, I kinda got somewhere, but I still have no idea what I'm doing or why it works (or doesn't).

Given $\lim_{n\to\infty} s_n = l$, determine $\sum_{k\ge0}(s_{k+2}-s_k)$

According to what I've seen on the site and further, we should be able to rearrange the terms to get

$$S_n = \sum_{n\ge0}(s_{n+2}-s_n)=-s_0-s_1+s_{n+1}+s_{n+2}$$

And taking the limit of which would lead us to

$$\lim_{n\to\infty}S_n = 2l-s_0-s_1$$

Now I may be picking bad examples, but I can't for the life of me figure out why that's regarded as correct. Since taking a sequence $\frac 1 {n+1} \to 0$, the sum should (by my logic) be $\sum_{n\ge0}\left(\frac 1 {n+3}-\frac 1 {n+1}\right) = 2 \times 0 - 1 - \frac 1 2$, which is $-\frac 3 2$. Yet, according to Wolfram, the true value of the sum is $\frac 5 6$, so surely there must be an error somewhere? Furthermore, I'm not even sure why would we even be allowed to use the commutative property here, since it doesn't hold unless the series converges absolutely, which is not specified in the problem. Does anyone mind shedding some light on all of this? Thank you!

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The idea is not to telescope the infinite series, but to telescope the partial sums. The $n$th partial sum is $S_n = -s_0-s_1+s_{n+2}+s_{n+1}$. Therefore the limit of the partial sums, $\lim_{n \to \infty} S_n$, is $$\lim_{n \to \infty} -s_0 - s_1 + s_{n+2} + s_{n+1} = -s_0 - s_1 + l + l = 2l-s_0-s_1$$

No commutativity in the limit required!

The true value of your sum is indeed $-\frac{3}{2}$ as you say; I don't know what you entered into WolframAlpha but it wasn't right.

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  • $\begingroup$ Whoops! My bad about the Wolfram thing, I forgot to set a lower bound and it assumed one. Yet, even with the correct boundaries, I get positive $\frac 3 2$, unless I'm doing something terribly wrong? Also, I'm not exactly sure of how we come to the expression for $S_n$. The way I would derive it is by expanding the sum, reordering a couple of terms, cancelling out the middle ones and what you said would remain. Yet, I'm unsure of whether we are allowed to simply reorder an infinite sum like that? $\endgroup$ – anonra Mar 28 '17 at 19:29
  • $\begingroup$ The different number is because you've entered into WA the negation of the sum you gave me. The expression for $S_n$ is a finite sum, not an infinite one, so its rearrangement is perfectly fine. $\endgroup$ – Patrick Stevens Mar 28 '17 at 21:09
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    $\begingroup$ Oh boy, I really messed that one up and confused myself more than I should have. Thank you for your help and apologies for kinda wasting your time on a bunch of silly typos. $\endgroup$ – anonra Mar 28 '17 at 21:51

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