If the series sum from n=1 to infinity a_n with a_n>0
If the series $\sum_{n=1}^{\infty} a_n$ with $a_n>0$ converges, prove that it can happen that $\varlimsup_{n\to\infty} \dfrac{a_{n+1}}{a_n}=q>1$.
Explanation of Method
The ratio test's hypothesis ($\limsup$ of consecutive ratios $<1$) is sufficient but not necessary for convergence — a series can converge even if the ratio spikes above $1$ infinitely often, as long as those spikes are rare and the overall terms still shrink fast enough. Constructing a sequence with occasional "jumps" at perfect squares, but still dominated overall by a convergent $p$-series, produces exactly this counterexample.
- Construct a sequence with a mild perturbation at perfect-square indices.
- Compute the ratio at the perturbed indices and show its limit exceeds $1$.
- Verify the resulting series still converges by comparison.
Step 1 · Construct the sequence
Define:
$$ a_n = \begin{cases} \dfrac2{n^2}, & n=k^2 \text{ for some integer } k \\[4pt] \dfrac1{n^2}, & \text{otherwise.} \end{cases} $$
Step 2 · Compute the ratio at perfect squares
At $n=k^2-1$ (just before the perfect square $k^2$):
$$ \frac{a_{k^2}}{a_{k^2-1}} = \frac{2/k^4}{1/(k^2-1)^2} = \frac{2(k^2-1)^2}{k^4} = 2-\frac4{k^2}+\frac2{k^4} \longrightarrow 2 \quad\text{as } k\to\infty. $$
So $\varlimsup_{n\to\infty}\dfrac{a_{n+1}}{a_n} \ge 2 > 1$.
Step 3 · Verify the series still converges
Since $a_n\le\dfrac2{n^2}$ for every $n$:
$$ \sum_{n=1}^{\infty} a_n \le \sum_{n=1}^{\infty}\frac2{n^2} = 2\cdot\frac{\pi^2}6 < \infty. $$
By the comparison test, $\sum a_n$ converges — despite the ratio exceeding $1$ infinitely often.
Summary Table
| Quantity | Value |
|---|---|
| $a_n$ at generic $n$ | $1/n^2$ |
| $a_n$ at $n=k^2$ | $2/n^2$ |
| $\varlimsup\, a_{n+1}/a_n$ | $\ge2>1$ |
| $\sum a_n$ | Converges (bounded by $2\sum1/n^2$) |
Additional Notes
- This construction cleanly separates the ratio test's sufficient condition from a necessary one: the series converges because every single term stays comparable to $1/n^2$, not because consecutive ratios stay below $1$ — the ratio test can fail to apply even when the series obviously converges by other means.
- The perturbation factor of $2$ at perfect squares is arbitrary; any bounded multiplicative spike, occurring at a sparse enough set of indices (like perfect squares, which have density $O(1/\sqrt n)$ among all integers), would work equally well.
- This example also shows that $\varliminf$ and $\varlimsup$ of the ratio can differ substantially (here $\varliminf=0$ since the ratio right after a spike, $\frac{a_{k^2+1}}{a_{k^2}}=\frac{1/(k^2+1)^2}{2/k^4}\to\frac12$, while other ratios tend to $1$), reinforcing why the ordinary limit-based ratio test can be too blunt an instrument for such sequences.
Result
A convergent series can indeed have $\varlimsup\dfrac{a_{n+1}}{a_n}>1$, exhibited by: