When using the alternating series test, if the limit is not equal to $0$, am I allowed to assume that it's divergent?

$$\sum_{n=1}^\infty (-1)^n\frac{(3n-1)}{(2n+1)}$$

Let $b_n=\frac{(3n-1)}{(2n+1)}$.

$\lim_{x\to \infty } \frac{3n-1}{2n+1}=\frac{3}{2}\ne0$

  • $\begingroup$ Yes, it diverges. $\endgroup$ – hamam_Abdallah Nov 8 '16 at 18:36
  • 1
    $\begingroup$ It's more or less like your favorite series $-1+1-1+1-1+1-1+1-\cdots$ at least on its tail $\endgroup$ – user160738 Nov 8 '16 at 18:37
  • $\begingroup$ In your series $\lim (-1)^nb_n$ does not exist. So it is non zero. $\endgroup$ – Bumblebee Nov 8 '16 at 18:49

You don't have to assume: it is trivially divergent since the general term of a convergent series tends to $0$.

Indeed if the series $S_n\sum_k=1^n u_k$ tends to $S$, for any $\varepsilon >0$, there's a rank $N$ such that $\lvert S_n-S\rvert<\varepsilon$ for all $n\ge N$. In particular, if $n> N$, $$\lvert u_n\rvert=\lvert S_n-S_{n-1}\rvert\le\lvert S_n-S\rvert+\lvert S-S_{n-1}\rvert <2\varepsilon.$$


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