# Alternating Series Test for Divergence

Alternating Series Test

if the alternating series $$\sum^\infty_{n=1}(-1)^{n-1}b_n=b_1-b_2+b_3-b_4+b_5+\dotsb+b_n\gt0$$ satisfies $$(\text{i)}\;\;b_{n+1}\leq b_n\;\;\;\;\;\text{for all } n,$$ $$(\text{ii)}\;\;\lim_{n\rightarrow \infty}b_n=0$$ then the series is convergent

Although the Alternating Series Test doesn't test for divergence, what if the series doesn't satisfy the second requirement (test fails)?

In this case, can we conclude it's divergence via Test for Divergence? (i.e. the limit doesn't exist)

Yes. If $\lim_{n\rightarrow\infty}b_n$ does not converge to $0$, then $\sum_{n=1}^{\infty}b_n$ does not exist - regardless of whether the series is alternating or not. In particular, if you define the series of partial sums as $$B_n=\sum_{i=1}^{n}b_i$$ then the sum $\sum_{n=1}^{\infty}b_n$ is defined as $\lim_{n\rightarrow\infty}B_n$. This implies that there is an $L$ such that, for any $\varepsilon$ and large enough $N$, it holds that for all $n>N$ that $B_n\in (L-\varepsilon,L+\varepsilon)$. This, in turn, implies that, since $B_{n+1}-B_{n}=b_{n+1}$ and since the maximum difference of two elements in $(L-\varepsilon,L+\varepsilon)$ is $2\varepsilon$, that $|b_{n+1}|<2\varepsilon$ for all $n>N$. This implies $b_n$ converges to $0$. The contrapositive of this statement is that if $b_n$ does not converge to $0$, the series does not converge.