The root test states the following:
Consider a series $\sum_{n = 0}^\infty a_n$ and denote $L := \limsup_{n\to\infty} \sqrt[n]{|a_n|}$. If $L < 1$, then the series is absolutely convergent, and if $L > 1$, the series is divergent.
However, I am interested in the finer question concerning the "badness" of the divergent case: For $L > 1$, can I comment anything on the boundedness of the sums $\big|\sum_{n = 0}^k a_n\big|$ as $k\to\infty$?
All I can seem to say is that we'll have a subsequence $(a_{N_k})$ for which the sums $\big|\sum_{n = 0}^k a_{N_k}\big|$ will be unbounded as $k\to\infty$. (Choose $a_{N_k} > L - \epsilon > 1$.)
If $\sum\limits_{k=1}^{n} a_k$ is bounded then so is $(a_n)$ because $a_n=\sum\limits_{k=1}^{n} a_k-\sum\limits_{k=1}^{n-1} a_k$. But if $L>1$ and we pick $C$ such that $1<C <L$ then $|a_n| >C^{n}$ for inifnitely many $n$ so $ (a_n)$ cannot be bounded.