Limit Comparison with SUM ln(n)/n^3 (Convergent or Divergent Series 23)

[@fariahuq3281]

isn't it if c>0 not c is greater or equal to 0?

[@matrixtrace]

Isn't it easier to just show that ln(n)/n^3 < n/n^3 = 1/n^2 and therefore converges?

[@nerky9663]

We had the same thought, I have a test where I had to test this series two different ways, and the first one I did was the integral test then I did direct comparison

[@thearsham]

URE A LIFE SAVER!!

[@carlosandresmarinjimenez7498]

No entiendo nada de inglés, pero vi el video y le entendí perfectamente

[@saigegilroy7751]

Thank you!

[@mathproblemssolvedcraigfau775]

My pleasure, Saige!

[@kavinyker6837]

Wow amazing tutor!

[@mathproblemssolvedcraigfau775]

Thanks Kaveet :)

[@sidneybourne9806]

BAD

[@quirklesteins2741]

agreed.

[@samanthamanoloff3580]

agree

[@samanthamanoloff3580]

calculus dave said so

[@sophiashakir8367]

yeeha obvi they don't understand that c has to equal literally ANYTHING BUT 0 or infinity smh