# Show that a series converges absolutely given convergence of two related series

#### tomcruisex

Show that if $$\displaystyle \sum_{n=1}^{\infty}a_n^2$$ and $$\displaystyle \sum_{n=1}^{\infty}b_n^2$$ are convergent then $$\displaystyle \sum_{n=1}^{\infty}a_nb_n$$ is absolutely convergent.

Hint: Show |xy| <= 1/2(|x|^2 + |y|^2) for any x,y ε R

I can prove the Hint -
$$\displaystyle |xy|\leq 1/2(|x|^2 + |y|^2)$$
$$\displaystyle 2|xy| \leq |x|^2 + |y|^2$$
$$\displaystyle |x|^2 + |y|^2 - 2|xy| \geq 0$$
$$\displaystyle (|x| - |y|)^2 \geq 0$$

And this is true for all x, y as the square of any real number will be greater than or equal to 0.

But I cant see how to apply this to the original question?

#### Prove It

MHF Helper
Show that if $$\displaystyle \sum_{n=1}^{\infty}a_n^2$$ and $$\displaystyle \sum_{n=1}^{\infty}b_n^2$$ are convergent then $$\displaystyle \sum_{n=1}^{\infty}a_nb_n$$ is absolutely convergent.

Hint: Show |xy| <= 1/2(|x|^2 + |y|^2) for any x,y ε R

I can prove the Hint -
$$\displaystyle |xy|\leq 1/2(|x|^2 + |y|^2)$$
$$\displaystyle 2|xy| \leq |x|^2 + |y|^2$$
$$\displaystyle |x|^2 + |y|^2 - 2|xy| \geq 0$$
$$\displaystyle (|x| - |y|)^2 \geq 0$$

And this is true for all x, y as the square of any real number will be greater than or equal to 0.

But I cant see how to apply this to the original question?
First notice that $$\displaystyle \displaystyle |x|^2 = x^2$$ and $$\displaystyle \displaystyle |y|^2 = y^2$$.

Since you have shown that $$\displaystyle \displaystyle |xy| \leq \frac{1}{2}\left(x^2 + y^2\right) = \frac{1}{2}x^2 + \frac{1}{2}y^2$$, that means

$$\displaystyle \displaystyle \sum{|xy|} \leq \frac{1}{2}\sum{x^2} + \frac{1}{2}\sum{y^2}$$.

Since you know that the two sums on the right are convergent, their sum is also convergent, and anything less than that is also convergent.

Q.E.D.

tomcruisex

#### chisigma

MHF Hall of Honor
Show that if $$\displaystyle \sum_{n=1}^{\infty}a_n^2$$ and $$\displaystyle \sum_{n=1}^{\infty}b_n^2$$ are convergent then $$\displaystyle \sum_{n=1}^{\infty}a_nb_n$$ is absolutely convergent.

Hint: Show |xy| <= 1/2(|x|^2 + |y|^2) for any x,y ε R

I can prove the Hint -
$$\displaystyle |xy|\leq 1/2(|x|^2 + |y|^2)$$
$$\displaystyle 2|xy| \leq |x|^2 + |y|^2$$
$$\displaystyle |x|^2 + |y|^2 - 2|xy| \geq 0$$
$$\displaystyle (|x| - |y|)^2 \geq 0$$

And this is true for all x, y as the square of any real number will be greater than or equal to 0.

But I cant see how to apply this to the original question?
Setting $$\displaystyle c_{n}^{2}= \text{max}\ [a_{n}^{2}\ ,\ b_{n}^{2}]$$ it is easy to demonstrate that the series $$\displaystyle \sum_{n=1}^{\infty} c_{n}^{2}$$ converges. But is $$\displaystyle |a_{n}\ b_{n}|\le c_{n}^{2}$$ so that the series $$\displaystyle \sum_{n=1}^{\infty} |a_{n}\ b_{n}|$$ also converges...

Kind regards

$$\displaystyle \chi$$ $$\displaystyle \sigma$$

tomcruisex

#### chisigma

MHF Hall of Honor
Hint: Show |xy| <= 1/2(|x|^2 + |y|^2) for any x,y ε R...
If $$\displaystyle |x\ y|\le \frac{|x|^{2} + |y|^{2}}{2}$$, then ...

$$\displaystyle x^{2}\ y^{2} \le \frac{x^{4}+2 x^{2} y^{2} + y^{4}}{4} \implies \frac{x^{2} y^{2}}{2} \le \frac{x^{4}+ y^{4}}{4} \implies x^{4}+y^{4} \ge 2 x^{2} y^{2}$$ (1)

Very well!... but what does it happen if $$\displaystyle x=y=\frac{1}{100}$$?...

Kind regards

$$\displaystyle \chi$$ $$\displaystyle \sigma$$

#### tomcruisex

Thank you for the help guys.

There is one thing I am still not sure about - what is the meaning of $$\displaystyle \sum_{n=1}^{\infty}a_n^2$$?
Does that mean that the series is made up by squaring each term and adding them together - a0a0 + a1a1 + a2a2 + ... + anan?
Or does it mean that the series you get the sum of series $$\displaystyle \sum_{n=1}^{\infty}a_n$$ and then square it?

#### chisigma

MHF Hall of Honor
Thank you for the help guys.

There is one thing I am still not sure about - what is the meaning of $$\displaystyle \sum_{n=1}^{\infty}a_n^2$$?
Does that mean that the series is made up by squaring each term and adding them together - a0a0 + a1a1 + a2a2 + ... + anan?
Or does it mean that the series you get the sum of series $$\displaystyle \sum_{n=1}^{\infty}a_n$$ and then square it?
The meaning of $$\displaystyle \sum_{n=1}^{\infty} a^{2}_{n}$$ is [probably...] that You have a series with all positive terms. The reason of that is easily understandable considering the case $$\displaystyle a_{n}=b_{n}= \frac{(-1)^{n}}{\sqrt{n}}$$, where both the series $$\displaystyle \sum_{n=1}^{\infty}a_{n}$$ and $$\displaystyle \sum_{n=1}^{\infty}b_{n}$$ converge but the series $$\displaystyle \sum_{n=1}^{\infty}a_{n} b_{n}$$ diverges...

Kind regards

$$\displaystyle \chi$$ $$\displaystyle \sigma$$