Question

Prove that if $\sum _{k=1}^{\infty }{a}_k$is a convergent series with $a_k\ge 0$for every positive integer k, then the series $\sum _{k=1}^{\infty }{a}_k^2$converges.

Short answer

Series $\sum _{k=1}^{\infty }{a}_k$is a convergent where $a_k\ge 0$for every positive integer k.

so $N>0$will also exist such that $a_k<1$for all $n>N.$

when $n>N\&a_k<1,$then$0\le a_k^2\le a_k.$

According to The Comparison Test, if $0\le a_k^2\le a_k\&\sum _{k=1}^{\infty }{a}_k$converges then series $\sum _{k=1}^{\infty }{a}_k^2$also converges.

Step-by-step solution

Step 1. Given Information.

Series$\sum _{k=1}^{\infty }{a}_k$is a convergent where$a_k\ge 0$for every positive integer k.

The given series are the following.

$\sum _{k=1}^{\infty }{a}_k^2$

Step 2. Proof.

Series $\sum _{k=1}^{\infty }{a}_k$is a convergent where $a_k\ge 0$for every positive integer k.

so $N>0$will also exist such that $a_k<1$for all role="math" localid="1649091708218" $n>N.$

when role="math" localid="1649092400735" $n>N\&a_k<1,$then $0\le a_k^2\le a_k.$

According to The Comparison Test, if role="math" localid="1649092225717" $0\le a_k^2\le a_k\&\sum _{k=1}^{\infty }{a}_k$converges then series role="math" localid="1649092217355" $\sum _{k=1}^{\infty }{a}_k^2$also converges.