Summation Woes

Well, here's something that both JX and Luoning can understand and appreciate; it's basically something I discussed with Shaun over MSN, and it's an alternating series, taking the form of:


Notice that we can group the terms like this and obtain a sum for an odd number of terms:


Also, we can group the terms like this and obtain a sum for an even number of terms:


The strange thing is, notice that if you consider an even number of terms, the sum goes to positive infinity, and if you consider an odd number of terms, the sum goes to negative infinity! This is what we call an oscillating series, which happens to be diverging as well.

I’ve included a graph for your reference, to illustrate the oscillating and diverging nature of this series:


Notice the diverging nature of the sum to n terms, and consequently, there can be no sum to infinity, simply because the sum to n terms depends on the very number of terms, and thus on the last term. A sum to infinity where the number of terms and the last term is not defined can therefore produce no well defined result.

Arithmetic This!

There’s a special name for this type of sequence:


Notice that each term differs from its preceding and succeeding term by a constant difference of 3 – this type of series can be represented as such:


Where a is the first term, and d is the common difference. Notice that also that the n-th term is given by a + (n – 1)d. This is known as an Arithmetic Progression, or an Arithmetic Sum.

Now, can we find a sum to n terms? That means, can we find a general formula for the following summation of the first term to the last term:


Well, of course we can! Let us see how; first, let us consider the sum to n terms on its own:


We started off by writing the sum from the first term a – but we could have done so by starting from the last term A as well, and thus the sum to n terms written backwards from the last term
A is simply:


Now let’s put them side by side and add them up and let’s see if you notice something:


What do you notice? Well, notice that the ‘d’s and the ‘-d’s all cancel one another out, leaving us with:


The n is there because we have n terms in each sum, and adding up two sums should give us n of A and n of a. Now, what is the last term A in terms of the first term a? We’ve already said that we want the sum to n terms, and thus the last term A must be the n-th term, given by:


In which case we make a substitution:


And therefore the sum to n terms follows nicely:

Sweet eh? :)