That too long!

For any natural number n > 1,we have :

(n - 1)n(n + 1) = n(n2 - 1) = n3 - n < n3

⇒1n3<1(n−1)n(n+1)

1(n−1)n(n+1)=1n.1(n−1)(n+1)

=1n.(n+1)−(n−1)(n−1)(n+1).12=12.1n.(1n−1−1n+1)

=12.(1(n−1)n−1n(n+1))

Now we have :

E < 12.3.4+13.4.5+14.5.6+...+1(n−1)n(n+1)

=12(12.3−13.4)+12(13.4−14.5)+12(14.5−15.6)+...+12(1(n−1)n−1n(n+1))

=12(12.3−1n(n+1))=112−12n(n+1)<112

Hence,E<112

For any natural number n > 1,we have :

(n - 1)n(n + 1) = n(n² - 1) = n³ - n < n³

\(\Rightarrow\dfrac{1}{n^3}< \dfrac{1}{\left(n-1\right)n\left(n+1\right)}\)

\(\dfrac{1}{\left(n-1\right)n\left(n+1\right)}=\dfrac{1}{n}.\dfrac{1}{\left(n-1\right)\left(n+1\right)}\)

\(=\dfrac{1}{n}.\dfrac{\left(n+1\right)-\left(n-1\right)}{\left(n-1\right)\left(n+1\right)}.\dfrac{1}{2}=\dfrac{1}{2}.\dfrac{1}{n}.\left(\dfrac{1}{n-1}-\dfrac{1}{n+1}\right)\)

\(=\dfrac{1}{2}.\left(\dfrac{1}{\left(n-1\right)n}-\dfrac{1}{n\left(n+1\right)}\right)\)

Now we have :

E < \(\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+\dfrac{1}{4.5.6}+...+\dfrac{1}{\left(n-1\right)n\left(n+1\right)}\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2.3}-\dfrac{1}{3.4}\right)+\dfrac{1}{2}\left(\dfrac{1}{3.4}-\dfrac{1}{4.5}\right)+\dfrac{1}{2}\left(\dfrac{1}{4.5}-\dfrac{1}{5.6}\right)+...+\dfrac{1}{2}\left(\dfrac{1}{\left(n-1\right)n}-\dfrac{1}{n\left(n+1\right)}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2.3}-\dfrac{1}{n\left(n+1\right)}\right)=\dfrac{1}{12}-\dfrac{1}{2n\left(n+1\right)}< \dfrac{1}{12}\)

Hence,\(E< \dfrac{1}{12}\)