# A sum

#### Integrator

Why?

I can't see how you could justify the term $$\displaystyle \frac{n + 1}{2}$$ in your first equation in such a way that it would still be justified in the second equation.
Hello,

I think I need to find another reasoning ... How do you think the proposed amount should be calculated?Thank you very much!

All the best,

Integrator

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#### skipjack

Forum Staff
If "amount" is a reference to the $$\displaystyle \frac{n + 1}{2}$$ term in your first equation, you didn't "propose" this term or give any reasoning for it - it seems to have been chosen to make the first equation work.

Changing the sign of the integral by exchanging its limits suggests that the sign of the $$\displaystyle \frac{n + 1}{2}$$ term should be changed as well, but then the second equation would be correct only if one decided that $$\displaystyle \sum_{k=n}^{k=1} k$$ means $$\displaystyle -\sum_{k=1}^{k=n} k$$, which would appear to be an arbitrary convention that doesn't make the equations more useful or easier to write or use.

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#### topsquark

Math Team
It is true that the calculation of a sum is somewhat like the calculation of an integral? I say yes and then we can write that $$\displaystyle \sum_{k=1}^{k=n} k=\frac{n+1}{2}+\int_1^nx dx$$
Where did you get this? Why is the integral there? For the record,
$$\displaystyle \sum_{k = 1}^n k = 1 + 2 + \text{ ... } + n = n + (n - 1) + \text{ ... } + 2 + 1 = \sum_{k = 1}^{n} (n - k + 1)$$

Both of these are equal to $$\displaystyle \dfrac{n(n + 1)}{2}$$, which can be easily shown by induction.

If you are clever enough you can sometimes find a way to sum the series that involves integration techniques, but this is not the case in general so far as I know. (But you can always use summation techniques to calculate an integral.)

-Dan

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#### Micrm@ss

A sum is just an integral over the counting measure. That is,
$$\sum_{k=1}^n a_k = \int_1^n a(k) d\mu,$$
where $\mu$ is the counting measure. In this point of view, you'd define
$$\sum_{k=1}^n a_k = -\sum_{k=n}^1 a_k.$$

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Forum Staff

#### idontknow

Something is going wrong with the bounds : $$\displaystyle \int_{a}^{n} =-\int_{n}^{a}$$ , which is being treated as a sum.

What I know is : $$\displaystyle n+[n-1]+[n-2]+...+1=n^2 -[1+2+...+n]=n^{2}-\frac{n(n+1)}{2}=\frac{n(n-1)}{2}.$$

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#### skipjack

Forum Staff
What you thought you knew should have been written as shown below:
$$\displaystyle n+[n-1]+[n-2]\,+\,...+\,1=n^2 -[1+2+\,...+\,(n-1)]=n^{2}-\frac{n(n-1)}{2}=\frac{n(n+1)}{2}$$.

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#### tahirimanov19

Sounds like one of Trump's opening lines ... You are assuming Trump is smart enough to have an opening line...

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#### Integrator

Hello all,

It is true that the simple integral is a sum of small surfaces and the proposed sum is numeric , but if $$\displaystyle \int_1^n x dx+\int_n^1 x dx=0$$ , then how much do I do $$\displaystyle \sum_{k=1}^{k=n} k+\sum_{k=n}^{k=1} k$$?Some say that $$\displaystyle \sum_{k=1}^{k=n} k+\sum_{k=n}^{k=1} k =n+1$$ and I don't understand this result!?

All the best,

Integrator

#### topsquark

Math Team
Some say that $$\displaystyle \sum_{k=1}^{k=n} k+\sum_{k=n}^{k=1} k =n+1$$ and I don't understand this result!?
Well, "some" are wrong as you can easily check for yourself!
$$\displaystyle \sum_{k = 1}^5 k + \sum_{k = 1}^5 (5 - k + 1) = (1 + 2 + 3 + 4 + 5) + (5 + 4 + 3 + 2 + 1) = 30 \neq 5 + 1$$

-Dan

Addendum: Oh! I see what's happening. Notice that
$$\displaystyle \sum_{k = 1}^n k + \sum_{k = 1}^n (n - k + 1) = \sum_{k = 1}^n ( k + n - k + 1) = \sum_{k = 1}^n (n + 1)$$

$$\displaystyle \sum_{k = 1}^n k + \sum_{k = 1}^n (n - k + 1) = \sum_{k = 1}^n (n + 1)$$

So the statement you are confused about is written wrong... you are missing the summation on the RHS.

You really need to start writing $$\displaystyle \sum_{k = n}^{k = 1} k$$ as the more standard $$\displaystyle \sum_{k = 1}^n (n - k + 1)$$ which is the clue you need.

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