normal distribution family tight

If we have $N(\mu_{n},\sigma^2_{n})$ distributions. I want to show that this family is tight if and only if the sequences $(\mu_{n})$ and $(\sigma^2_{n})$ are bounded.

A sequence of probability measures $(\nu_{n})$ on $(\mathbb{R},B)$ is called tight, if $$\lim_{M\to \infty}\inf_n \ \nu_{n}([-M,M])=1.$$

I'd like to use characteristic functions but i cannot figure out how to do it. Could anyone help me with this?

Also under what conditions do the $N(\mu_{n},\sigma^2_{n})$ distributions converge to a (weak) limit? And what limit is that?

I'd really appreciate some help on this.

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Tightness is a necessary and sufficient condition to have weak convergence of a subsequence. From that, we can argue by contradiction. –  Davide Giraudo Nov 27 '12 at 21:24
Ok so we have a subsequence that weakly converges to a limit? –  Math Girl Nov 27 '12 at 21:49
Yes. So in particular the characteristic function converges pointwise. –  Davide Giraudo Nov 27 '12 at 21:50
Characteristic functions are not needed here, simply using the definition of tightness you recall will do. (And please get rid of the confusing $\mu_n$ notation.) –  Did Nov 27 '12 at 22:13
Please do not vandalize your post. –  Did Nov 27 '12 at 23:22
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