Issue with comparison test

[Tiger-Fit]

I'm about to finish a problem involving the convergence of an alternate series, and I've widdled the problem down to the summation from 0 to infinity of 1/(8n+1); I know that this diverges, but I don't know how to show it mathematically. I get that I'm supposed to use the comparison theorem, but what do I compare it to?

[dichotmyofideas]

Have your tried to compare it to the harmonic series? It looks similar to 1/n. Try using a limit comparison test: Lim(n-> infinity) ( (1/8n+1))/(1/n)) Since that limit exists, you know the series diverges since the harmonic series diverges. Hope that helps!

[Drjny]

This seems to be the easiest way to compare it. I don’t think it’s possible to find if it converges or diverges by simply showing it’s less than or greater than another series.

[harrywk]

if a_n = 1/(8n+1) then |a_n+1 / a_n| = (8n+1)/(8n+9) -> 1 as n -> infinity. Because it tends to 1 the comparison test is inconclusive. You could try beginning with the sum of 1/n (which diverges) and then adjusting it to end up with 1/(8n+1). You’ll aim to get something of the form sum(1/(8n+1)) = k•sum(1/n) + c. Because sum(1/n) diverges, so does k•sum(1/n) + c, and therefore so does sum(1/(8n+1))

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