Sum of an infinitive series, with variable being a natural number

Question

I have the following problem

Let $n$ be a natural number. What is the sum of $1+2+3+...+(n-1)+n$

I'm not 100% certain what would be a correct way of calculating it. What I've done so far is look at the first and second first and last term. $1+n$ and $2+(n-1)$.

So I have $2n-1+2+3+...$. So that means the sum is $∞$

@Christian Notice that $1+n$, $2+(n-1)$, $3+(n-2)$ and so on each add up to $n+1$. Since there are $(n+1)-1$ terms from $1$ to $n$, and each pair has two terms, the sum is equal to $(n+1)(\frac{n}{2})$. When $n$ tends to infinity the sum also tends to infinity. — Toby Mak, Aug 26 '17 at 12:52
@MichaelRozenberg He asks: What is the sum of $1+2+3+\cdots+(n-1)+n$ — José Carlos Santos, Aug 26 '17 at 12:54
@JoséCarlosSantos In the title he asks for the sum when it is an infinite series — Toby Mak, Aug 26 '17 at 12:55

score 1 · Answer 1 · answered Aug 26 '17 at 12:51

1

HINT: the solution by Gauss: $$1+2+3+4+...+97+98+99+100=(100+1)+(99+2)+(98+3)+(97+4)+...$$

answered Aug 26 '17 at 12:51

Dr. Sonnhard Graubner

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Sum of an infinitive series, with variable being a natural number

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