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 $\sum_{n=1}^{\infty} a_n^2$ also converges. Show by example that the converse is not true.
Explanation of Method
A necessary condition for $\sum a_n$ to converge is $a_n\to0$, so eventually every term is less than $1$ — and squaring a number less than $1$ only makes it smaller. That inequality $a_n^2<a_n$ (for large $n$) lets the comparison test transfer convergence from $\sum a_n$ down to $\sum a_n^2$. The converse fails because squaring can push a divergent series into convergence, as $a_n=1/n$ demonstrates.
- Use the necessary condition $a_n\to0$ to show $a_n<1$ eventually.
- Apply the comparison test using $a_n^2<a_n$.
- Exhibit a counterexample for the converse using $a_n=1/n$.
Part 1 · Proof that $\sum a_n$ convergent $\Rightarrow$ $\sum a_n^2$ convergent
Since $\sum a_n$ converges, $\lim_{n\to\infty}a_n=0$, so there exists $N$ such that $a_n<1$ for all $n\ge N$. For such $n$, since $0<a_n<1$:
$$ a_n^2 < a_n. $$
Since $\sum_{n=N}^{\infty} a_n$ converges (a tail of a convergent series), by the comparison test $\sum_{n=N}^{\infty} a_n^2$ converges too. Adding back the finite sum $\sum_{n=1}^{N-1}a_n^2$:
$$ \sum_{n=1}^{\infty} a_n^2 = \sum_{n=1}^{N-1}a_n^2 + \sum_{n=N}^{\infty}a_n^2 < \infty. $$
Part 2 · Counterexample for the converse
Let $a_n=\dfrac1n$. Then $\sum a_n^2=\sum\dfrac1{n^2}$ converges ($p=2>1$, sum $=\frac{\pi^2}6$), but $\sum a_n=\sum\dfrac1n$ is the harmonic series, which diverges. So $\sum a_n^2$ converging does not imply $\sum a_n$ converges.
Summary Table
| Direction | Status |
|---|---|
| $\sum a_n$ conv. $\Rightarrow$ $\sum a_n^2$ conv. | True (proved via comparison test) |
| $\sum a_n^2$ conv. $\Rightarrow$ $\sum a_n$ conv. | False ($a_n=1/n$ counterexample) |
Additional Notes
- The inequality $a_n^2<a_n$ for $0<a_n<1$ is the entire mechanism at work: squaring a fraction between $0$ and $1$ always shrinks it, so convergent series (whose terms must eventually fall below $1$) automatically produce even-faster-converging squared series.
- The failure of the converse illustrates that squaring can convert borderline-divergent decay rates (like $1/n$, order $1$) into comfortably-convergent ones (like $1/n^2$, order $2$) — the operation is not reversible in general.
- This result and its converse are foundational to the theory of $\ell^2$ (square-summable) sequences: $\ell^1\subset\ell^2$ for positive sequences, since absolute summability is a strictly stronger condition than square-summability.
Result
The forward implication holds; the converse fails: