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 $\sum _{k=1}^{\infty }\frac{a_k}{k}$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$0\le \frac{a_k^{}}{k}\le a_k.$

According to The Comparison Test, if role="math" localid="1649139323101" $0\le \frac{a_k^{}}{k}\le a_k\&\sum _{k=1}^{\infty }{a}_k$converges then series $\sum _{k=1}^{\infty }\frac{a_k^{}}{k}$also converges.

Step-by-step solution

Step 1. Given Information. 

The given series is$\sum _{k=1}^{\infty }\frac{a_k}{k}.$

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

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 $n>N\&a_k<1,$thenrole="math" localid="1649139290641" $0\le \frac{a_k^{}}{k}\le a_k.$

According to The Comparison Test, if role="math" localid="1649139270963" $0\le \frac{a_k^{}}{k}\le a_k\&\sum _{k=1}^{\infty }{a}_k$converges then series $\sum _{k=1}^{\infty }\frac{a_k^{}}{k}$also converges.