# Arithmetic of series

1. Jun 15, 2005

### steven187

hello all

a simple question how come

$$\sum_{k=1}^{n} k- \sum_{k=1}^{n} (k-1) \not= 1$$

even though i know if you expand it out you would get n, isnt there an arithmetic property of series that relates to this?

thanxs

2. Jun 15, 2005

### whozum

The sum of the consecutive sequence of integers starting with 1 is

$$\frac{1}{2}n(n+1)$$ where n is the last (and largest) member of the sequence.

so from here we can see that the set of any consecutive set of integers has the sum given by

$$\frac{1}{2}n_1(n_1+1) - \frac{1}{2}n_2(n_2+1)$$, combining:

$$\frac{1}{2}(n_1^2 + n_1 - n_2^2 - n_2)$$

The difference of your two series lies in the fact that the first element of the second series is just 0 whereas the first series has 1. The difference between the two sums will always be 'n', which can be shown by realizing that in the above expression the relationship between n1 and n2 is just $n_2 = n_1 - 1 [/tex]. Last edited: Jun 15, 2005 3. Jun 15, 2005 ### steven187 hello there see i know how to do it through this method $$\sum_{k=1}^{n} k- \sum_{k=1}^{n} (k-1) = \frac{n(n+1)}{2}-\frac{n(n-1)}{2}=n$$ but what i want to know is why cant we do this step? $$\sum_{k=1}^{n} k- \sum_{k=1}^{n} (k-1)= \sum_{k=1}^{n} k-k+1= \sum_{k=1}^{n} 1=1$$ is it that we are not allowed to do subtraction between 2 series, if not are there any cases when we can? thanxs Last edited: Jun 15, 2005 4. Jun 15, 2005 ### master_coda Of course you can do subtraction between series. But [itex]\sum_{k=1}^n1=n$ and not 1.

5. Jun 16, 2005

### steven187

hello there

thanxs for that, i should have remembered something simple like that

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