I need to calculate this:
$\sum_{k=1}^{n}(\frac{1}{k}-\frac{1}{k+1})$
The answear is:
$1 - \frac{1}{n+1}$
But I have no idea how to get to that answear. I tried to simplify the equation to: $\frac{1}{kx^n+k}$ But now I don't know If I should put it inside the formula for geometric sums or if it's even possible.
Thank you!
$$\sum_{k=1}^{n}(\frac{1}{k}-\frac{1}{k+1})=\sum_{k=1}^{n}\frac{1}{k}-\sum_{k=1}^{n}\frac{1}{k+1}=\sum_{k=0}^{n-1}\frac{1}{k+1}-\sum_{k=1}^{n}\frac{1}{k+1}$$ $$=1+\sum_{k=1}^{n-1}\frac{1}{k+1}-\sum_{k=1}^{n}\frac{1}{k+1}$$
Interestingly, the formula does not simplify when you reduce to the common denominator:
$$\sum_{k=1}^n\frac1{k(k+1)}$$
doesn't look easier.
By if you expand a few terms using the original form, the solution all of a sudden looks trivial.
