examples using comparison test without limit

Do the following series converge?

${\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{\displaystyle \frac{1}{n^2+n+1}}$ (1)

${\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{\displaystyle \frac{n^3+n+1}{n^4+n+1}}$ (2)

The general of (1) may be estimated upwards:

Because  ${\sum }_{n=1}^{\mathrm{\infty }}\frac{1}{n^2}$ (an over-harmonic series) converges, then also (1) converges.

The general of (2) may be estimated downwards:

$$\frac{n^3+n+1}{n^4+n+1}>\frac{n^3+0+0}{n^4+n^4+n^4}=\frac{1}{3}\cdot \frac{1}{n}>0$$

Because ${\sum }_{n=1}^{\mathrm{\infty }}\frac{1}{3}\frac{1}{n}$ (the harmonic series divided by 3) diverges, then also (2) diverges.

Title	examples using comparison test without limit
Canonical name	ExamplesUsingComparisonTestWithoutLimit
Date of creation	2013-03-22 15:08:55
Last modified on	2013-03-22 15:08:55
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	9
Author	pahio (2872)
Entry type	Example
Classification	msc 40-00
Related topic	PTest
Defines	over-harmonic series