# 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 $\frac{1}{4}$.

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