Try rationalizing the numerator.analysis001 said:Homework Statement
Prove that the series [itex]\sum_{n=1}^{\infty}\frac{\sqrt{n+1}-\sqrt{n}}{n}[/itex] converges.
The Attempt at a Solution
I think I'm going to use the comparison test but I'm having trouble coming up with a series to compare it to. Any clues would be great. Thanks!
SammyS said:Try rationalizing the numerator.
Let's see ...analysis001 said:Yeah, I've gotten to that point, so as of now I have: [itex]\sum_{n=1}^{\infty}\frac{1}{n(\sqrt{n+1}+\sqrt{n})}[/itex] but I'm still not sure what to compare it to.
The given series is \sum_{n=1}^{\infty}\frac{\sqrt{n+1}-\sqrt{n}}{n}.
To prove the convergence of a series, one can use various methods such as the comparison test, ratio test, root test, integral test, or the limit comparison test. These tests involve comparing the given series to a known convergent or divergent series and using the properties of the known series to determine the convergence of the given series.
The comparison test is a method used to prove the convergence or divergence of a series by comparing it to a known series. If the given series is less than or equal to a known convergent series, then the given series is also convergent. If the given series is greater than or equal to a known divergent series, then the given series is also divergent.
To apply the comparison test to the given series, we can compare it to the series \sum_{n=1}^{\infty}\frac{1}{n}. Since \frac{\sqrt{n+1}-\sqrt{n}}{n} is always less than or equal to \frac{1}{n}, we can conclude that the given series is convergent as it is bounded by a known convergent series.
Yes, the given series can also be proven to be convergent using the root test or the integral test. However, the comparison test is the most efficient method in this case as it requires the least amount of work.