I´m sorry I still don´t get how to get the answer

So I have the same issue. I posted 1.5hours ago.

Thank you hallsofIvy but I don´t grasp how to continue.

Sigma from k=1 to n $\displaystyle (1/k-1/(k+1))$

How on earth do I get that to be $\displaystyle 1-1/(1+n)$?

I tried this $\displaystyle 1/2+1/6+1/12 +1/(n-1) -1/n$. Then I try to get every term that includes an n to the same denominator. so I get $\displaystyle n/n(n-1) -(n-1)/n(n-1)$

so I get $\displaystyle (n-n+1)/n(n-1) = 1/(n^2-n)$ and ofcourse that is almost very very not right.

So well that didn´t work.

Well I tried the other way $\displaystyle 1/2+1/6....+1/n-1/(n+1)$ that gives me $\displaystyle 1/(n^2+n)$

And that doesn´t help....I really don´t need a complete solution maybe a hint would work wonders....