Given that sum from n=1 to infinity a_n² and sum from n=1
Given that $\sum_{n=1}^{\infty} a_n^2$ and $\sum_{n=1}^{\infty} b_n^2$ converge, prove that the following series converge: $$ 1^\circ\; \sum_{n=1}^{\infty}|a_nb_n|; \qquad 2^\circ\; \sum_{n=1}^{\infty}(a_n\pm b_n)^2; \qquad 3^\circ\; \sum_{n=1}^{\infty}\frac{|a_n|}{n}. $$
Explanation of Method
All three results are consequences of the AM-GM and Cauchy-Schwarz inequalities, which bound a product or weighted sum by the square-summable quantities already assumed convergent. Once $\sum|a_nb_n|$ converges, expanding $(a_n\pm b_n)^2$ shows it's a sum of three already-convergent series; and Cauchy-Schwarz with the known-convergent $\sum1/n^2$ handles the third part directly.
- Bound $|a_nb_n|$ using AM-GM, and sum to get part $1^\circ$.
- Expand $(a_n\pm b_n)^2$ and use part $1^\circ$ to get part $2^\circ$.
- Apply Cauchy-Schwarz with $\sum1/n^2$ to get part $3^\circ$.
Part $1^\circ$ · $\sum |a_nb_n|$ converges
By AM-GM, $|a_nb_n|=\sqrt{a_n^2}\sqrt{b_n^2}\le\dfrac{a_n^2+b_n^2}2$. Summing:
$$ \sum_{n=1}^{\infty}|a_nb_n| \le \frac12\left(\sum_{n=1}^{\infty}a_n^2+\sum_{n=1}^{\infty}b_n^2\right) < \infty. $$
By the comparison test, $\sum|a_nb_n|$ converges. (Equivalently, Cauchy-Schwarz gives $\sum_{n=1}^N|a_nb_n|\le\sqrt{\sum a_n^2}\sqrt{\sum b_n^2}$, a bounded partial sum.)
Part $2^\circ$ · $\sum (a_n\pm b_n)^2$ converges
Expanding: $(a_n\pm b_n)^2 = a_n^2 \pm 2a_nb_n + b_n^2$. Summing:
$$ \sum_{n=1}^{\infty}(a_n\pm b_n)^2 = \sum a_n^2 \pm 2\sum a_nb_n + \sum b_n^2. $$
Each piece converges: $\sum a_n^2$, $\sum b_n^2$ by hypothesis, and $\sum a_nb_n$ converges absolutely by Part $1^\circ$. A sum of convergent series converges, so $\sum(a_n\pm b_n)^2$ converges.
Part $3^\circ$ · $\sum \dfrac{|a_n|}{n}$ converges
By Cauchy-Schwarz applied to $|a_n|$ and $\frac1n$:
$$ \left(\sum_{n=1}^{N}\frac{|a_n|}{n}\right)^2 \le \left(\sum_{n=1}^{N}a_n^2\right)\left(\sum_{n=1}^{N}\frac1{n^2}\right). $$
Since $\sum a_n^2$ converges (given) and $\sum\frac1{n^2}$ converges (to $\frac{\pi^2}6$), the right side is bounded as $N\to\infty$. So the partial sums of $\sum\frac{|a_n|}n$ are bounded and increasing, hence convergent.
Summary Table
| Part | Key tool |
|---|---|
| $1^\circ$ | AM-GM: $|a_nb_n|\le\frac12(a_n^2+b_n^2)$ |
| $2^\circ$ | Expansion + Part $1^\circ$ |
| $3^\circ$ | Cauchy-Schwarz with $\sum1/n^2$ |
Additional Notes
- These are the series-theoretic analogues of standard facts about $\ell^2$ (square-summable) sequence spaces: the results here show that $\ell^2$ is closed under products (Cauchy-Schwarz/Hölder), closed under addition (the parallelogram-type expansion), and embeds into a weighted $\ell^1$ space via multiplication by $1/n$.
- Part $2^\circ$'s proof genuinely depends on Part $1^\circ$: without knowing $\sum a_nb_n$ converges, the cross term in the expansion could not be handled, illustrating how these results build on each other rather than standing independently.
- Part $3^\circ$ generalizes readily: replacing $\frac1n$ with any sequence $c_n$ satisfying $\sum c_n^2<\infty$ (not just $c_n=1/n$) gives the same conclusion by the identical Cauchy-Schwarz argument.
Result
All three series converge: