First I noticed that 1/8 can be factored out of each term, as follows:
1/8 * (1/1 + 1/3 + 1/6 + 1/10 + 1/15 + 1/21 + ...)
Notice an interesting pattern with the denominators inside the parentheses?
From the first term to the second term, the difference between the denominators = 2
From the second term to the third term, the difference between the denominators = 3
From the third term to the fourth term, the difference between the denominators = 4
..and so on
It took some trial-and-error on my part, but the terms within the parentheses can be expressed as the sum of all terms given by 1/((n^2 + n)/2), as 'n' goes from 1 to infinity.
For example, when n=1, 1/((1^2 + 1)/2) = 1/1
...when n=2, 1/((2^2 + 2)/2) = 1/3
...when n=3, 1/((3^2 + 3)/2) = 1/6
...when n=4, 1/((4^2 + 4)/2) = 1/10, etc.
Also, note that 1/((n^2 + n)/2) can be expressed as 2/(n^2 + n)
Now, the entire sum = 1/8 * (sum of all terms given by 2/(n^2 + n), as 'n' goes from 1 to infinity)
The '2' in the numerator can be factored out as well, resulting in:
1/4 * (sum of all terms given by 1/(n^2 + n), as 'n' goes from 1 to infinity)
However, the expression 1/(n^2 + n) = 1/((n)(n+1)) = 1/n - 1/(n+1)
When expressed as 1/n - 1/(n+1), the terms become (1-1/2) + (1/2 - 1/3) + (1/3 - 1/4), etc...
In the end, all the terms are cancelled out, except for the '1' at the beginning.
So, the overall sum is now 1/4 (which was the constant taken outside the series) * 1 = 1/4.
I'm really old, but I'll never be too old to become more educated.
