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) I am new so sorry if my noobness shines through.
In calculus I thought we learned that when faced with polynomial over polynomial like this, because the powers are equal on top and bottom this should converge to the ratio of the first coefficients which makes sense because they have the highest impact due to the x^2. Is this only true for limits of functions and not series here?
In an analysis book I am working through, there is a term test that says if a series converges then lim (an) is 0. However this seems contradictory because that implies the series I am working on diverges because the limit is non zero?
Which way of thinking is correct here? Does this series converge to 1/3 or diverge from the term test?
 A: 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.$
