If the series sum from n=1 to infinity a_n converges
If the series $\sum_{n=1}^{\infty} a_n$ converges and $a_n>a_{n+1}>0$ for sufficiently large $n$, prove that: $$ \lim_{n\to\infty} n a_n = 0. $$
Explanation of Method
The Cauchy criterion says the tail sum $\sum_{k=n+1}^{2n}a_k$ shrinks to $0$; since the sequence is decreasing, that tail sum — made of $n$ terms — is at least $n$ times its smallest member, $a_{2n}$. Squeezing these two facts together forces $n\,a_{2n}\to0$, and a short argument extends this from the even subsequence to the full sequence.
- Use the Cauchy criterion to bound the tail sum from $n+1$ to $2n$.
- Use monotonicity to bound that same tail sum below by $n\cdot a_{2n}$.
- Combine both bounds and extend from $a_{2n}$ to the general term $a_n$.
Step 1 · Cauchy criterion bound
Since $\sum a_n$ converges, for every $\varepsilon>0$ there is $N_0$ such that for $n\ge N_0$ (and using $a_k>0$):
$$ \sum_{k=n+1}^{2n} a_k < \varepsilon. $$
Step 2 · Monotonicity bound
Since $a_n$ is eventually decreasing, every term $a_k$ for $k$ in $\{n+1,\ldots,2n\}$ satisfies $a_k\ge a_{2n}$ (the smallest one in that range). There are exactly $n$ such terms:
$$ n\cdot a_{2n} \le \sum_{k=n+1}^{2n} a_k. $$
Step 3 · Combine and extend to the general term
Combining Steps 1 and 2: $n\cdot a_{2n} < \varepsilon$ for $n\ge N_0$, so $\lim_{n\to\infty} n\,a_{2n}=0$, equivalently $\lim_{m\to\infty}(m/2)a_m=0$ along even $m=2n$. For general $n$, using monotonicity: if $2m\le n<2(m+1)$, then $a_n\le a_{2m}$, so:
$$ n\,a_n \le 2(m+1)a_{2m} = 2m\,a_{2m} + 2a_{2m}. $$
Since $2m\,a_{2m}\to0$ (from the even-subsequence result, doubled) and $a_{2m}\to0$ (necessary condition for convergence), the right side $\to0$, so $n\,a_n\to0$ by the squeeze principle (as $n\,a_n\ge0$ always).
Summary Table
| Step | Result |
|---|---|
| Cauchy criterion | $\sum_{k=n+1}^{2n}a_k\to0$ |
| Monotonicity | $n\,a_{2n}\le\sum_{k=n+1}^{2n}a_k$ |
| Combined | $n\,a_{2n}\to0$, extended to $n\,a_n\to0$ |
Additional Notes
- Monotonicity of $a_n$ is essential here — without it, the lower bound $n\,a_{2n}\le\sum_{k=n+1}^{2n}a_k$ fails, since a non-monotone sequence could have $a_{2n}$ much larger than other terms in the range while the tail sum still shrinks.
- A standard counterexample when monotonicity is dropped: $a_n=\frac{(-1)^n}n$ isn't positive, but a genuinely non-monotone positive analogue can be built where $\sum a_n$ converges yet $na_n$ does not tend to $0$, showing the hypothesis is not merely a convenience.
- This result is the discrete companion to the fact that if $\int_1^\infty f(x)\,dx$ converges with $f$ positive and decreasing, then $xf(x)\to0$ — both are manifestations of the same "tail mass must vanish faster than the index grows" principle.
Result
For any convergent series with eventually decreasing positive terms: