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$\sum_{k=1}^\infty a_k$ converges and $a_k > 0$. Prove that $\sum_{k=1}^\infty \frac{a_k}{3 + a_k}$ also converges using the comparison test.

I figure that $3 + a_k > a_k$, so $a_k > \frac{a_k}{3 + a_k}$. In other words, dividing by something larger than the numerator will always make the series element smaller, therefore, by the comparison test, if the larger series converges, then the smaller series also converges.

Please help me understand the flaw in my reasoning and how I could do it better. Thank you!

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  • $\begingroup$ i was marked down for it on an exam. the professor isnt available for another week or so, but its been on my mind since getting the results back. looking to get as much information as i can in the mean time. thank you. $\endgroup$ – tau Jul 23 at 23:44
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    $\begingroup$ $$3+a_k > a_k \implies 1 > \frac{a_k}{3+a_k}.$$ $\endgroup$ – Dzoooks Jul 23 at 23:49
  • $\begingroup$ i see how that implication works. could i also say $3+a_k > 1 \implies \frac{a_k}{3+a_k} < \frac{a_k}{1}$? thank you for the help! $\endgroup$ – tau Jul 24 at 0:03
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    $\begingroup$ @tau Yes, if you had said "$3+a_k>1$, so $a_k>\frac{a_k}{3+a_k}$" your reasoning would be correct. $\endgroup$ – grand_chat Jul 24 at 0:10
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    $\begingroup$ doh, i see my mistake. thank you for the help. it took me awhile to realize that my premise was all wrong. $\endgroup$ – tau Jul 24 at 1:12
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$3+a_k >a_k$ does not give you $a_k >\frac {a_k} {3+a_k}$. We have $\frac {a_k} {3+a_k} <\frac {a_k} 3$. Now apply comparison test.

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  • $\begingroup$ can you please explain more? i am misunderstanding something basic... i do not see how my implication is false. thank you! $\endgroup$ – tau Jul 23 at 23:55
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    $\begingroup$ See the comment above by Dzoooks. $\endgroup$ – Kavi Rama Murthy Jul 24 at 0:02

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