Let (xn)n≥1 be a non-negative real sequence, that is, xn ≥ 0 for all n ≥ 1. Supposethat the series P xn converges. Prove that the seriesX x2nalso converges.Hint) The convergence of P xn implies limn→∞ xn = 0. Consequently, (xn)n≥1 should bea bounded sequence.

To prove that the series ∑x²n also converges, we can use the comparison test for series convergence.

Step 1: Given that the series ∑xn converges, we know that lim(n→∞) xn = 0.

Step 2: Since xn is a non-negative real sequence, we know that xn ≥ 0 for all n ≥ 1.

Step 3: Because lim(n→∞) xn = 0, there exists some N such that for all n > N, xn < 1.

Step 4: For these n > N, we have 0 ≤ x²n ≤ xn (because 0 ≤ xn < 1).

Step 5: Since ∑xn converges and 0 ≤ x²n ≤ xn for n > N, by the comparison test, ∑x²n also converges.

Therefore, if the series ∑xn converges, then the series ∑x²n also converges.