Q. 53 - Calculus

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.

Step-by-Step Solution

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 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 $n>N.$

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.