Let $(a_n)_{n=m}^{\infty}$ be a sequence of nonzero real numbers.
Then we know that the implication
$$\limsup\limits_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|>1\Rightarrow \sum_{n=m}^{\infty}a_n\ \text{ diverges}$$
is false
(a counterexample is the sequence $(a_n)_{n\geq 1}:=\begin{cases}\frac{1}{n^2} & \text{if } n \text{ is even }\\ \frac{2}{n^2} & \text{if }n \text{ odd}\end{cases}$ which has $\limsup\limits_{n\to\infty} |\frac{a_{n+1}}{a_n}|=2>1$ but nonetheless converges (as can be easily seen by using Comparison Test with the series $\sum_{n=1}^{\infty}\frac{2}{n^2}$).
This notwithstanding I haven't been able to see where the following proof of the false statement, which I had come up with before stumbling upon the counterexample I described above, is wrong, so I would appreciate if someone could point it out to me. Thanks.
Let $L=\limsup\limits_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right|$; then there exists a subsequence $\left(b_k\right)_{k\in\mathbb{N}}=\left(\left|\frac{a_{n_k+1}}{a_{n_k}}\right|\right)_{k\in\mathbb{N}}$ such that $\lim\limits_{k\to\infty}b_k=L$ so if we take $\varepsilon:=\frac{L-1}{2}$ we have that there exists $K\in\mathbb{N}$ such that $|b_k-L|\leq\varepsilon$ for every $k\geq K$ hence $b_k\geq\frac{L+1}{2}>1$ for every $k\geq K.$ So if $k>K$ we have $|a_{n_k}|=\left|\frac{a_{n_k}}{a_{n_{k-1}}}\right|\cdot \left|\frac{a_{n_{k-1}}}{a_{n_{k-2}}}\right| \cdot \dots \cdot \left|\frac{a_{n_{K+1}}}{a_{n_K}} \right|\cdot |a_{n_K}|>\left(\frac{L+1}{2}\right)^{k-K}|a_{n_K}|=A\left(\frac{L+1}{2}\right)^k$, where $A:=\left(\frac{L+1}{2}\right)^{-K}|a_{n_K}|$ which implies that $\lim\limits_{k\to\infty}|a_{n_k}|\geq\lim\limits_{k\to\infty} A\left(\frac{L+1}{2}\right)^k=+\infty$ i.e. $\lim\limits_{k\to\infty}|a_{n_k}|=+\infty$ hence $\lim\limits_{k\to\infty} |a_{n_k}|\neq 0$ thus $\lim\limits_{n\to\infty}a_n\neq 0$ and therefore the series $\sum\limits_{n=0}^{\infty}a_n$ cannot converge, as desired. $\square$
Best Answer
The problem is after you split it up into a product. Indeed while it is true that
$$\lvert a_{n_k}\rvert=\left\lvert\frac{a_{n_k}}{a_{n_{k-1}}}\right\rvert\cdots\left\lvert\frac{a_{n_{K+1}}}{a_{n_{K}}}\right\rvert\cdot\lvert a_{n_K}\rvert,$$
you have to observe that the factors are on the form $\left\lvert\frac{a_{n_{j+1}}}{a_{n_{j}}}\right\rvert$ and not on the form $\left\lvert\frac{a_{n_j+1}}{a_{n_{j}}}\right\rvert=b_j$. It is very important where the shift in the indexing occurs, and it is this issue which causes the argument to fail, as the inequality cannot actually be applied.