Summation on HP 42S

[Albert Chan]

(09-26-2018 12:36 PM)Gerson W. Barbosa Wrote:  As I am kinda lazy, I’ll just ask my HP 50g to do it for me:

'∑(k=1,n,2/(k*(k+3)))'
EVAL
COLLECT

What does it say ? I bet it look complicated ...

Actually, the sum is very easy, once you recognized above formula

∑(k=1,n,2/(k*(k+3)))
= 2/3 * ∑(k=1,n,3/(k*(k+3)))
= 2/3 * ∑(k=1,n, 1/k - 1/(k+3))

If you tried out different values of k, the sum have at most 6 terms. All the rest cancelled out.

∑(k=1,n,2/(k*(k+3))) = 2/3 * (1 + 1/2 + 1/3 - 1/(n+1) - 1/(n+2) - 1/(n+3))