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Discovered something interesting while I was playing around.

Let $c$ be the sum of all naturals, i.e. $c = 0 + 1 + 2 + 3 + \ldots$

Then $3c = 3 + 6 + 9 + 12 + \ldots$

It appears that every term in $3c$ is equal to the sum of three immediately following terms in $c$. e.g.

$$3 = 0 + 1 + 2$$ $$6 = 1 + 2 + 3$$ $$9 = 2 + 3 + 4$$ etc. So basically $$3c = (c_0 + c_1 + c_2) + (c_1 + c_2 + c_3) + (c_2 + c_3 + c_4) + \ldots $$ where $c_i$ is the $i$th term of $c$. Why is this? It looks related to the partial sums of $c$, e.g. 1, 3, 6, 10, 15, etc. Probably missing something obvious, but that's the way it goes.

Thanks for any help in figuring this out.

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    Well, you are manipulating a series that doesn't exist in the normal sense. – 2017-01-28
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    How do you define "the sum of all naturals" ? – 2017-01-28
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    @SimplyBeautifulArt I know that it diverges, but the phenomena is there even if you only sum up to $n$. – 2017-01-28
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    Clearly $3n=(n-1)+n+(n+1)$,hence the pattern. – 2017-01-28
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    Damn. So I was missing something obvious. Cheers, kingW3. – 2017-01-28
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    @InstantRamen It's still not useful to think of it as a series rather than merely sequence. – 2017-01-28

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We have that $3n=(n-1)+n+(n+1)$ from that you can write each term as sum of $3$ consecutive terms.