Question

determine if series converges or diverges using the comparison test:
sum(n=1 to infinity)[3/(8n+7sqrt(n))]

Answer 1

diverges for sure

Answer 2

you can compare it to
\[\sum\frac{1}{5n}\]

Answer 3

since \(\sqrt{n}<n\)

Answer 4

clear? or no?

Answer 5

no, not really..

Answer 6

ok lets take it one step at a time

Answer 7

\(\sqrt{n}<n\) for any \(n\) so \(8n+7\sqrt{n}<8n+7n=15n\)

Answer 8

since \(15n>8n+7\sqrt{n}\) taking reciprocals tells you
\[\frac{3}{8n+7\sqrt{n}}>\frac{3}{15n}\]

Answer 9

\[\sum\frac{3}{15n}=\frac{1}{5}\sum\frac{1}{n}\] the well known divergent harmonic series
since \(\sum\frac{1}{n}\) diverges and since your series has larger terms, it too must diverge by the comparison test

Answer 10

Thanks A Lot!!! :)

Answer 11

yw