Convergence from another series 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? 
 A: 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.
A: $$0 \le (x - y)^2 = x^2 - 2xy + y^2$$
$$xy \le \frac 12 x^2 + \frac 12 y^2$$
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.
