MHB Facebook's question regarding a sum

[alyafey22]

I asked the following question on facebook

Prove that

$$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4} \, \cdot\cdot \cdot \,+ \frac{1}{n(n+1)}=\frac{n}{n+1}
$$

 

[alyafey22]

Prove that

$$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4} \, \cdot\cdot \cdot \,+ \frac{1}{n(n+1)}=\frac{n}{n+1}
$$

1- By inspection through partial sums

We know that the sum can be written as

$$\sum^{n}_{k=1}\frac{1}{k(k+1)}$$

Let $$S_{n}=\sum^{n}_{k=1}\frac{1}{k(k+1)}$$

$$S_1 = \frac{1}{2}$$

$$S_2=\frac{1}{2}+\frac{1}{6}=\frac{2}{3}$$

$$S_3=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}=\frac{3}{4}$$

$$S_4=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}=\frac{4}{5}$$

$$S_n=\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4} \, \cdot\cdot \cdot \,+ \frac{1}{n(n+1)}=\frac{n}{n+1}$$

---

2- The second approach is using telescoping series

$$\sum^{n}_{k=1}\frac{1}{k(k+1)}=\sum^{n}_{k=1}\frac{1}{k}-\frac{1}{k+1}$$

$$\sum^{n}_{k=1}\frac{1}{k}-\frac{1}{k+1}=\left( 1 - \frac{1}{2} \right) +\left( \frac{1}{2} - \frac{1}{3} \right) +\left( \frac{1}{3} - \frac{1}{4} \right)+\cdot \cdot \cdot + \left(\frac{1}{n} - \frac{1}{n+1} \right)$$

Cancelling out the terms we reach to

$$\sum^{n}_{k=1}\frac{1}{k}-\frac{1}{k+1}= 1 -\frac{1}{n+1} =\frac{n}{n+1}$$

 

[MarkFL]

Very nice!

Your second approach is what I would use, partial fraction decomposition on the summand and then observing the result is a telescopic series.

In the first approach, I would require a student to then complete the hypothesis via induction.(Happy)