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Facebook's question regarding a sum

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
I asked the following question on facebook

Prove that

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

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Prove that

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

1- By inspection through partial sums

We know that the sum can be written as

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

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

\(\displaystyle S_1 = \frac{1}{2}\)

\(\displaystyle S_2=\frac{1}{2}+\frac{1}{6}=\frac{2}{3}\)

\(\displaystyle S_3=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}=\frac{3}{4}\)

\(\displaystyle S_4=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}=\frac{4}{5}\)

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

[HR][/HR]

2- The second approach is using telescoping series

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

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

Cancelling out the terms we reach to

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

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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)