Anneliese H. answered • 11/03/21
Highly Trained Tutor for High School and College Level Math
While not perfect, let's use this symbol for the product notation ∏
One note, I am assuming you meant the function to be 1/(k+1), where k+1 is on the bottom of the fraction. For the function 1/k + 1 it's much easier!
So we have ∏(1/(k+1)) from k=4 to n
So we would have:
∏(1/(k+1)) = 1/(4+1) × 1/(5+1) × 1/(6+1) × 1/(7+1) × ... × 1/(n-1 + 1) × 1/(n+1)
= 1/5 × 1/6 × 1/7 × 1/8 ×...×1/n-1 × 1/n × 1/(n+1)
= 1/(5×6×7×8×...×n) × 1/(n+1)
= 1/n! × 1/(n+1)
= 1/n!(n+1)
= 1/ ((n+1)! + n!)
Where all that beautiful business is on the bottom of the fraction.
If the function is 1/k + 1 where 1 is a constant added to the fraction 1/k, it is much easier! Following the same steps you would get 1/n! + 1 as you stated in your question.
Hope this helps!
Anneliese H.
Because I made a mistake! I was multiplying n! through, but you will get 1/(n!n + n!). So probably neater to leave it as 1/n!(n+1) anyway. Thanks for catching it!11/03/21
Pippi A.
Oh, ok. Thanks.11/03/21
Pippi A.
Would the problem change if the conditions on n was that "n is an integer greater than 10" or would it still be this answer?11/06/21
Anneliese H.
It shouldn't, as 10 is a positive integer. Issues could arise with integers from -1 and lower (the function is undefined at n=-1). Also, I realize now that n!(n+1)=(n+1)!11/06/21
Pippi A.
Thank you, how does the problem move from 1/n!(n+1) to the final answer of 1/ ((n+1)!+n!)?11/03/21