A sum

Aug 2018
137
7
România
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
Dec 2006
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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
May 2013
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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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Oct 2009
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367
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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Dec 2015
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Earth
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
Dec 2006
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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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Aug 2018
137
7
România
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
May 2013
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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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