Why does the term test not match the 'limit calculus trick of convergence' for the infinite series of (k^2 - 3k + 1) / (3k^2 + k - 2)

Solution 1:

Suppose $\ a_n\ $ is a sequence of real numbers with $\ a_n \not\to 0.\ $ Then $\ \sum\ a_n\ $ does not converge. I prove this result now.

By definition of "not converges to $0",\ \exists \varepsilon>0\ $ such that $\ \vert a_n \vert > \varepsilon\ $ for infinitely many integers $\ n.\ (*)$

Suppose $\ \displaystyle \lim_{k\to\infty}\left(\sum_{i=1}^{k} a_i\right)\ = L\in\mathbb{R}.\ $Then $ \exists\ N\in\mathbb{N}\ $ such that $\ \displaystyle \lvert L - \left(\sum_{i=1}^{k} a_i\right)\ \rvert < \frac{\varepsilon}{2}\ $ for all $ k\geq N.$

Furthermore, $\ (*)\implies\ $there exists $\ N'>N\ $ such that $\ \vert a_{N'}\vert > \varepsilon.\ $

However, since

$\displaystyle\lvert L - \left(\sum_{i=1}^{N'-1} a_i\right)\ \rvert < \frac{\varepsilon}{2},\ $ i.e. $\ L + \frac{\varepsilon}{2} >\displaystyle\left(\sum_{i=1}^{N'-1} a_i\right)\ > L - \frac{\varepsilon}{2},\ $ we must have either$ \ \displaystyle\left(\sum_{i=1}^{N'} a_i\right)\ = \left(\sum_{i=1}^{N'-1} a_i \right)+ a_{N'}\ > L + \frac{\varepsilon}{2}\ $ or $ \ \displaystyle\left(\sum_{i=1}^{N'} a_i\right)\ < L - \frac{\varepsilon}{2},\ $ contradicting the assumption that $\ \displaystyle \lvert L - \left(\sum_{i=1}^{k} a_i\right)\ \rvert < \frac{\varepsilon}{2}\ $ for all $ k\geq N.$