The problem is simply, determine whether the series converges or diverges. Thanks!
T thrownaway New member Joined Feb 1, 2021 Messages 3 Feb 1, 2021 #1 The problem is simply, determine whether the series converges or diverges. Thanks!
pka Elite Member Joined Jan 29, 2005 Messages 11,982 Feb 1, 2021 #2 Hint: 1log(n)+n>12n\large\dfrac{1}{\log(n)+\sqrt{n}}>\dfrac{1}{2\sqrt{n}}log(n)+n1>2n1 for n≥2n\ge 2n≥2
Hint: 1log(n)+n>12n\large\dfrac{1}{\log(n)+\sqrt{n}}>\dfrac{1}{2\sqrt{n}}log(n)+n1>2n1 for n≥2n\ge 2n≥2