This problem is the Exercise from Discrete Mathematics book, p.24. The problem says
Find the a closed form of $$S_n = \sum_{1 \leq k \leq n}\frac{1}{k(k + 1)}$$
I tried these ways with no luck
Re-index and plus themself
Replace k with k + 1
From $$S_n = \sum_{1 \leq k \leq n}\frac{1}{k(k + 1)}$$ Replace k with k + 1 $$S_n = \sum_{1 \leq k + 1 \leq n}\frac{1}{(k + 1)(k + 2)}$$ Add -1 all through the inequation $$S_n = \sum_{0 \leq k \leq n - 1}\frac{1}{(k + 1)(k + 2)}$$ The range is not matched with original. So, I think I cannot go further.
Replace k with k - 2
From $$S_n = \sum_{1 \leq k \leq n}\frac{1}{k(k + 1)}$$ Replace k with k - 2 $$S_n = \sum_{1 \leq k - 2 \leq n}\frac{1}{(k - 2)(k - 1)}$$ Add +2 all through the inequation $$S_n = \sum_{3 \leq k \leq n + 2}\frac{1}{(k - 2)(k - 1)}$$ The range is not matched with original. So, I think I cannot go further.
Replace k with n - k + 1
From $$S_n = \sum_{1 \leq k \leq n}\frac{1}{k(k + 1)}$$ Replace k with n - k + 1 $$S_n = \sum_{1 \leq n - k + 1 \leq n}\frac{1}{(n - k + 1)(n - k + 2)}$$ Tune the inequation $$S_n = \sum_{0 \leq n - k \leq n - 1}\frac{1}{(n - k + 1)(n - k + 2)}$$ $$S_n = \sum_{-n \leq -k \leq -1}\frac{1}{(n - k + 1)(n - k + 2)}$$ $$S_n = \sum_{1 \leq k \leq n}\frac{1}{(n - k + 1)(n - k + 2)}$$ Try adding them together $$2S_n = \sum_{1 \leq k \leq n}(\frac{1}{k(k + 1)} + \frac{1}{(n - k + 1)(n - k + 2)})$$ $$2S_n = \sum_{1 \leq k \leq n} \frac{(n - k + 1)(n - k + 2) + k(k + 1)} {k(k + 1)(n - k + 1)(n - k + 2)}$$ I decide not to go further. Is this the dead end?
Plus with extra boundary
From $$S_n = \sum_{1 \leq k \leq n}\frac{1}{k(k + 1)}$$ Add $$\frac{1}{(n + 1)(n + 2)}$$ to both side $$S_n + \frac{1}{(n + 1)(n + 2)} = \sum_{1 \leq k \leq n}\frac{1}{k(k + 1)} + \frac{1}{(n + 1)(n + 2)}$$ $$S_n + \frac{1}{(n + 1)(n + 2)} = \sum_{1 \leq k \leq n + 1}\frac{1}{k(k + 1)}$$ Adjust the inequation $$S_n + \frac{1}{(n + 1)(n + 2)} = \frac{1}{2} \sum_{1 \leq k - 1 \leq n}\frac{1}{k(k + 1)}$$ Replace k with k + 1 $$S_n + \frac{1}{(n + 1)(n + 2)} = \frac{1}{2} \sum_{1 \leq k \leq n}\frac{1}{(k + 1)(k + 2)}$$ From this point, I have no idea to go next.
Can anyone give me a clue? Every suggestion will be appreciated.
In general, try calculating this kind of thing for several small $n$. You might see a pattern. $$\{S_1,S_2,S_3,S_4,\ldots\}=\left\{\frac{1}{2},\frac{2}{3},\frac{3}{4},\frac{4}{5},\ldots\right\}$$ Does a pattern jump out? If you can propose a formula for $S_n$, you could try induction to prove that formula is correct.