Prove that $\sum\limits_{n=1}^{2000} \frac {1}{n^3+3n^2+2n}\lt \frac {1}{4}$

Question

Prove that $$\sum_{n=1}^{2000} \frac {1}{n^3+3n^2+2n}\lt \frac {1}{4}.$$

I don't know a way to prove this, just instantly saw that:

$$n^3+3n^2+2n = n(n+1)(n+2).$$

But think this isn't useful at all.

Any hints?

Answer 1

In fact

$ \displaystyle \frac{1}{n(n+1)(n+2)} = \dfrac 12 \dfrac{(n+2)-n}{n(n+1)(n+2)} = \dfrac12 \left( \dfrac{1}{n(n+1)} - \dfrac{1}{(n+1)(n+2)} \right)$

Answer 2

Hint: The factorization you cite should suggest a partial fraction decomposition leading to a telescoping series. This would prove that the sum all the way to infinity is exactly $\frac 14$. Since your sum deletes some terms, it is less.

Answer 3

Partial fractions: \begin{align*} \frac1{n^3+3n^2+2n}&=\frac1{n(n+1)(n+2)}\\ &=\frac{\frac12}{n}-\frac{1}{n+1}+\frac{\frac12}{n+2} \end{align*}

Then we can prove (by induction) $$\sum_{n=1}^M\frac1{n^3+3n^2+2n}=\frac12-\frac14-\frac1{2(M+1)}+\frac1{2(M+2)}=\frac14-\frac1{2(M+1)(M+2)}$$

Answer 4

one should have the following :

\begin{align} \sum_{n=1}^{2000}\frac{1}{n(n+1)(n+2)}&=\sum_{n=1}^{2000}\bigg(\frac{1}{n(n+1)}-\frac{1}{n(n+2)}\bigg)\\ &=\sum_{n=1}^{2000}\frac{1}{n(n+1)}-\sum_{n=1}^{2000}\frac{1}{n(n+2)}\\ &=\sum_{n=1}^{2000}\bigg(\frac{1}{n}-\frac{1}{n+1}\bigg)-\frac{1}{2}\sum_{n=1}^{2000}\bigg(\frac{1}{n}-\frac{1}{n+2}\bigg)\\ &=\bigg(1-\frac{1}{2001}\bigg)-\frac{1}{2}\bigg(1+\frac{1}{2}-\frac{1}{2001}-\frac{1}{2002}\bigg)\\ &=\frac{1}{4}-\frac{1}{2001}+\frac{1}{2}\cdot\frac{1}{2001}+\frac{1}{4004}\\ &=\frac{1}{4}-\frac{1}{4002}+\frac{1}{4004}\\ &<\frac{1}{4}-\frac{1}{4002}+\frac{1}{4002}\\ &=\frac{1}{4} \end{align}