# An Inequality With an Infinite Series

The following problem has been given in the 1970s at the oral entrance examination at the Mathematics Department (MechMat) of the Moscow State University. (Several of the Department's alumni have been collecting sample problems, of a rather unexpected difficulty. The one at hand (#15) is among the easiest in the collection.

Prove that

 (1) $\displaystyle \sum_{n=1}^{1000}\frac{1}{n^{3}+3n^{2}+2n} \lt \frac{1}{4}$

It can be easily verified (or obtained by the method of the undetermined coefficients) that

 (2) $\displaystyle \frac{1}{n^{3}+3n^{2}+2n} = \frac{1}{2}\bigg(\frac{1}{n}-\frac{2}{n+1}+\frac{1}{n+2}\bigg)$

which naturally points to the telescoping property of the series.

We shall show that the inequality holds with $1000$ replaced by $N \gt 0$.

$\displaystyle \sum_{n=1}^{N}\frac{1}{n + 1} = \sum_{n=1}^{N}\frac{1}{n} - 1 + \frac{1}{N + 1}$ and
$\displaystyle \sum_{n=1}^{N}\frac{1}{n + 2} = \sum_{n=1}^{N}\frac{1}{n} - 1 -\frac{1}{2}+ \frac{1}{N + 1}+ \frac{1}{N + 2}.$

Using these we may continue with (1) and (2) thus

\displaystyle \begin{align} \sum_{n=1}^{N}\frac{1}{n^{3}+3n^{2}+2n} &= \frac{1}{2}\sum_{n=1}^{N}\bigg(\frac{1}{n}-\frac{2}{n+1}+\frac{1}{n+2}\bigg) \\ &= \frac{1}{2}\bigg(\sum_{n=1}^{N}\frac{1}{n} \\ &- 2\bigg(\sum_{n=1}^{N}\frac{1}{n} - 1 + \frac{1}{N + 1}\bigg) \\ &+ \sum_{n=1}^{N}\frac{1}{n} - 1 -\frac{1}{2}+ \frac{1}{N + 1}+ \frac{1}{N + 2}\bigg) \\ &= \frac{1}{2}\bigg(2-\frac{2}{N + 1} - 1 -\frac{1}{2}+ \frac{1}{N + 1}+ \frac{1}{N + 2}\bigg) \\ &= \frac{1}{2}\bigg(\frac{1}{2}-\frac{1}{N+1}+\frac{1}{N+2}\bigg) \\ &= \frac{1}{4}-\frac{1}{2(N+1)(N+2)} \lt \frac{1}{4}. \end{align}

Obviously, the series in (1) is convergent with the sum of $\frac{1}{4}$. Its partial sums are increasing and bounded from above by $\frac{1}{4}$.

### Telescoping situations

• Leibniz and Pascal Triangles
• Infinite Sums and Products
• Sum of an infinite series
• Harmonic Series And Its Parts
• A Telescoping Series
• That Divergent Harmonic Series
• An Elementary Proof for Euler's Series
• $\sin 1^{\circ}+\sin {2^\circ}+\sin 3^{\circ}+\cdots+\sin 180^{\circ}=\tan 89.5^{\circ}$
• Problem 3824 from Crux Mathematicorum
• $x_n=\sin 1+\sin 3+\sin 5+\cdots+\sin (2n-1)$
• Dan Sitaru's Sum of a Series
• A Property of Product of Special Matrices, Probably Folklore