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Suppose that the series $\displaystyle\sum_{n=1}^{\infty}b^2_{n}$ of postive numbers diverges. Prove that there exists a sequence $\{a_{n}\}$ of real numbers such that $$ \sum_{n=1}^{\infty}a^2_{n}<\infty \quad\text{and}\quad \sum_{n=1}^{\infty}|a_{n}b_{n}|=\infty. $$

My try: maybe this Cauchy-Schwarz inequality have usefull $$\Big(\sum_{n=1}^{\infty}a^2_{n}\Big)\Big(\sum_{n=1}^{\infty}b^2_{n}\Big)\ge \Big(\sum_{n=1}^{\infty}a_{n}b_{n}\Big)^2$$

Martin Argerami
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math110
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2 Answers2

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Define $$ s_n=\sum_{k=1}^nb_k^2 $$ and $$ a_n=\frac{b_n}{\sqrt{s_ns_{n-1}}} $$ Without loss of generality, assume $b_1\ne0$.

Since $u-1\ge\log(u)$ for $u\gt0$, $$ \begin{align} \sum_{k=2}^n a_kb_k &=\sum_{k=2}^n\frac{s_k-s_{k-1}}{\sqrt{s_ks_{k-1}}}\\ &=\sum_{k=2}^n\sqrt{\frac{s_k}{s_{k-1}}}-\sqrt{\frac{s_{k-1}}{s_k}}\\ &\ge\sum_{k=2}^n\sqrt{\frac{s_k}{s_{k-1}}}-1\\ &\ge\frac12\sum_{k=2}^n\log\left(\frac{s_k}{s_{k-1}}\right)\\[8pt] &=\frac12(\log(s_n)-\log(s_1)) \end{align} $$ Therefore, $$ \sum_{k=1}^\infty a_kb_k=\infty $$

$$ \begin{align} \sum_{k=2}^n a_k^2 &=\sum_{k=2}^n\frac{s_k-s_{k-1}}{s_ks_{k-1}}\\ &=\sum_{k=2}^n\frac1{s_{k-1}}-\frac1{s_k}\\ &=\frac1{s_1}-\frac1{s_n} \end{align} $$ Therefore, $$ \sum_{k=2}^\infty a_k^2=\frac1{b_1^2} $$

robjohn
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Let $S_N= \sum_{n=1}^N b_n^2$ and take $a_n=b_n/\sqrt{S_N}$. Then

$$\sum_{n=1}^N a_n^2 = 1$$

and

$$\sum_{n=1}^N a_n b_n = \sqrt{S_N}$$.

Let $N\to\infty$.

JPi
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    How does this lead to the sequence $a_n$? If we take the limit as $N\to\infty$, we get $a_n=0$. – robjohn Jan 22 '14 at 21:58
  • The question was to establish its existence, not necessarily what it looked like. – JPi Jan 22 '14 at 22:43
  • What sequence has its existence established here? All of these sequences are tending to the sequence $a_n=0$. – robjohn Jan 22 '14 at 22:57
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    Put another way: For every finite $N$, your infinite sequence $(a_n)$ yields $\sum_{n=1}^\infty a_n^2 = \infty$, violating the required condition. – ccorn Jan 22 '14 at 22:59