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Suppose that both $$\sum_{j=0}^n a_j^2$$ and $$\sum_{j=0}^n b_j^2$$ are convergent. Show that $$\sum_{j=0}^n a_jb_j$$ converges absolutely.

Ok, so my final exam is tomorrow and I have been working on this problem trying to figure it out a while. Can someone please help me?

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    $\begingroup$ $0\le(a-b)^2=a^2+b^2-2ab$. $\endgroup$ – David Mitra Apr 30 '16 at 6:40
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Hint: Look up Cauchy-Schwarz inequality.

Edit: By the Cauchy-Schwarz inequality, we have $$ \sum_{i=1}^n |a_ib_i|\le \sqrt{\sum_{i=1}^n a_i^2}\sqrt{\sum_{i=1}^n b_i^2}\ . $$

By your assumption, $\sqrt{\sum_{i=1}^n a_i^2}$ and $\sqrt{\sum_{i=1}^n b_i^2}$ converge so the let hand side also converges and the following relation holds: $$ \sum_{i=1}^{\infty} |a_ib_i|\le \sqrt{\sum_{i=1}^{\infty} a_i^2}\sqrt{\sum_{i=1}^{\infty} b_i^2}\ . $$

It should now be obvious that $\sum_{i=1}^{n} a_ib_i$ converges absolutely.

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  • $\begingroup$ I looked up the Cauchy-Schwarz and I can't really find how that correlate. Could you go into more depth please? $\endgroup$ – Dillon Malone Apr 30 '16 at 6:51
  • $\begingroup$ @DillonMalone Ok... I'll edit the answer to add more details. $\endgroup$ – BigbearZzz Apr 30 '16 at 6:54
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$$0 \le (x - y)^2 = x^2 - 2xy + y^2$$ $$xy \le \frac 12 x^2 + \frac 12 y^2$$ That is, $$|a_jb_j| \le \frac 12 |a_j|^2 + \frac 12 |b_j|^2 = \frac 12 a_j^2 + \frac 12 b_j^2$$ $$\sum_{k = 1}^{\infty} |a_jb_j| \le \frac 12 \sum_{k = 1}^{\infty} a_j^2 + \frac 12 \sum_{k = 1}^{\infty} b_j^2 \lt \infty$$

And we're done.

This can also be proved from Cauchy-Schwarz inequality, in a similar manner. On a side note, the Cauchy-Schwarz inequality may be derived from the inequality obtained above, using a technique known as normalisation. And $xy \le 1/2 x^2 + 1/2 y^2$ is a special case of the AM-GM (Arithmetic Mean-Geometric Mean) inequality.

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  • $\begingroup$ Thank you very much. This makes it more clear. $\endgroup$ – Dillon Malone May 1 '16 at 9:29

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