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Prove that the limit of an2 = 0 if the limit of an = 0

Prove that

    \[ \lim_{n \to  \infty} a_n^2 = 0 \]

if

    \[ \lim_{n \to \infty} a_n = 0. \]


Proof. Since

    \[ \lim_{n \to \infty} a_n = 0 \]

we know that for all \varepsilon > 0 there exists a positive integer N_1 such that

    \[ \lim_{n \to \infty} a_n = 0 \quad \implies \quad |a_n| < \varepsilon \]

for all n \geq N. In particular, if \varepsilon = 1 then there exists an integer N_2 such that

    \[ |a_n| < 1 \qquad \text{for all } n \geq N_2. \]

So, for any \varepsilon > 0 take N = \max \{ N_1, N_2 \} then we have

    \[ |a_n| < \varepsilon \qquad \text{and} \qquad |a_n| < 1 \]

for all n \geq N. (Since n \geq N implies n \geq N_1 and n \geq N_2.) Then, since |a_n| < 1 implies |a_n^2| < 1 we have

    \[ |a_n| < \varepsilon \quad \implies \quad |a_n^2| < \varepsilon \]

for all n \geq N. Hence,

    \[ \lim_{n \to \infty} |a_n^2| = 0. \qquad \blacksquare\]

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