Question

Series of squares Prove that if \(\sum a_{k}\) is a convergent series of positive terms, then the series \(\Sigma a_{k}^{2}\) also converges.

Short answer

Answer: Yes, if the series ∑ ak converges, then the series ∑ ak^2 also converges.

Step-by-step solution

Understand the given condition

The given condition is that the series \(\sum a_{k}\) converges, which means that the sum of an infinite number of terms of the series has a finite limit.

Comparison Test

To apply the Comparison Test, we need to show that the terms of the squared series \(\sum a_{k}^2\) are smaller (or equal) than the terms of the given convergent series \(\sum a_{k}\). Since the terms are positive, we can compare the ratio of consecutive terms: For \(a_k > 0\), let's consider the ratio of the consecutive terms of the series: \(\frac{a_{k}}{a_{k-1}}>1.\) Raise both sides of the inequality to the power of 2: \(\left(\frac{a_{k}}{a_{k-1}}\right)^2>1 \Rightarrow \frac{a_{k}^2}{a_{k-1}^2}>1.\) Now, multiply both sides of the inequality by \(a_{k-1}^2\) to obtain: \(a_k^2 > a_{k-1}^2.\) This means that the terms of the squared series are decreasing.

Monotonicity and Boundedness of Partial Sums of the Squared Series

Since the terms of the squared series are decreasing, we can say that partial sums of the squared series: \(S_n=a_1^2+a_2^2+\cdots+a_n^2\) are both monotonic (non-decreasing) and bounded from above (by the limit of the given convergent series). Therefore, by the Monotone Convergence Theorem, the sequence of partial sums converges, and thus the series \(\sum a_{k}^{2}\) converges.

Conclusion

Based on the Comparison Test, we have shown that if the given series \(\sum a_{k}\) converges, then the squared series \(\sum a_{k}^{2}\) also converges.