Question

Prove that if \({a_n} \ge 0\) and \(\sum {{a_n}} \)converges then \(\sum {{a^2}_n} \)also converges.

Short answer

If \({a_n} \ge 0\) and \(\sum {{a_n}} \)converges then \(\sum {{a^2}_n} \) also converges.

Step-by-step solution

Using property of convergence:-

We have \({a_n} \ge 0\)

And \(\sum {{a_n}} \)converges

Then, \(\mathop {\lim }\limits_{t \to \infty } {a_n} = 0\)

By definition of limit:-

So there exist a number N such that \(|{a_n} - 0| < 1\) for all \(n > N\)

\(0 \le {a_n} < 1\)for all \(n > N\)---(1)

Multiplying (1) by \({a_n}\):-

\(0 \le {a^2}_n < {a_n}\)

Since, \(\sum {{a_n}} \)converges so, by comparison test \(\sum {{a^2}_n} \) will be convergent.

Hence, \(\sum {{a^2}_n} \) will also be converges.