Vrftsjtryk

Let ${x_n}$ be a convergent sequence and ${y_n}$ a divergent sequence. Prove that ${ax_n + by_n}$ diverges for $bne 0$.

Intuitively this is obvious to me, however i want a formal proof.

Using the definition of a limit we know that:

$$
lim_{nto infty}ax_n = A stackrel{text{def}}{iff} {forall varepsilon > 0 exists N in mathbb N : forall n ge N implies |ax_n - A| < varepsilon}
$$

On the other hand we know that $y_n$ diverges, thus:
$$
lim_{n to infty}by_n = exists! stackrel{text{def}}{iff} {exists varepsilon >0 N in mathbb N : exists n ge N implies |by_n - B| ge varepsilon}
$$

Choose some $varepsilon$:
$$
begin{cases}
|ax_n - A| < {varepsilon over 2} \
|by_n - B| ge {varepsilon over 2}
end{cases}
$$
or:
$$
begin{cases}
|ax_n - A| < {varepsilon over 2} \
-|by_n - B| le -{varepsilon over 2}
end{cases}
$$

This is where I got stuck. It feels like i have to use some sort of triangular inequality (either direct or reverse) and find a contradiction. How do I proceed from this point?

There's more than one way to skin a cat.

The easiest way to prove your point is by knowing another fact, that is:

If ${a_n}$ converges and ${b_n}$ converges and $alpha, betainmathbb R$, then ${alpha a_n + beta b_n}$ also converges.

Using this, you can easily construct a proof by contradiction. That is, you can, assuming that ${x_n}$ converges and ${ax_n + by_n}$ converges, prove that ${y_n}$ must also converge, since $$y_n = frac{1}{b}left(ax_n + by_nright) + left(-frac{a}{b}right)x_n$$

However, if you insist on going by definitions, then you shouldn't just "choose some $epsilon$." The fact that ${y_n}$ does not converge gives you a place where you can start to build your $epsilon$. In particular, choosing some $B$, you can take the $epsilon_y$ which satisfies the property that $forall Ninmathbb Nexists n>N: |y_n - B|>epsilon$.

You can then use this to prove that $acdot A + bcdot B$ cannot be a limit, and since every number can be written as $aA+bB$ for some value $B$, this proves the sequence doesn't converge.

For the sake of completeness I'm putting here my further steps based on the hints.

Let:
$$
lim_{nto infty}x_n = A\
lim_{nto infty}y_n = B\
$$

We want to show that $lim_{ntoinfty}(ax_n + by_n)$ exists. Suppose $x_n$ and $y_n$ converge. It is necessary and sufficient $(1)$ to show that:

$$
x_n = A + alpha_n\
y_n = B + beta_n
$$

Where $alpha_n$ and $beta_n$ are infinitely small sequences. Then by definition of a limit we have:
$$
|x_n - A| < varepsilon iff |alpha_n| < varepsilon iff lim_{ntoinfty}alpha_n = 0 \
|y_n - B| < varepsilon iff |beta_n| < varepsilon iff lim_{ntoinfty}beta_n = 0
$$

Take some $a$ and $b$ and consider the following limit:
$$
lim_{ntoinfty}(ax_n + by_n) = lim_{ntoinfty}left((aA + bB) + (aalpha_n + bbeta_n)right) = aA + bB + lim_{ntoinfty}(aalpha_n + bbeta_n)
$$

But any linear combination of infinitely small sequences is an infinitely small sequence. So:

$$
lim_{ntoinfty}(ax_n + by_n) = aA + bB
$$

Now consider the case when $y_n$ diverges, that means that:
$$
|y_n - B| ge varepsilon iff |beta_n| ge varepsilon iff lim_{ntoinfty}beta_n = exists !
$$

So since $y_n$ may not be presented as a sum of some constant and and infinitely small sequence (which violates $(1)$) then $(ax_n + by_n)$ must also diverge.