Convergent/divergent series

[ludwig259]

Hey, i'd appreciate some help on finding for which real a does this series converge.
I've tried rewriting it with partial fractions and than using the comparison test, but it didn't seem to work out.

 

[blamocur]

For [imath]n >= 1[/imath]: [imath]\frac{1}{n} \leq \frac{1+n^2}{1+n^3} \leq \frac{2}{n}[/imath]

 

[Deleted member 4993]

Hey, i'd appreciate some help on finding for which real a does this series converge.
I've tried rewriting it with partial fractions and than using the comparison test, but it didn't seem to work out.

View attachment 30746


View attachment 30747

You know that \(\displaystyle \sum_{n=0}^\infty {\frac{1}{n}} \) is called a harmonic series and it it divergent.

 

[ludwig259]

You know that \(\displaystyle \sum_{n=0}^\infty {\frac{1}{n}} \) is called a harmonic series and it it divergent.

True, I sadly can't see how this could help solve the problem.

 

[AvgStudent]

True, I sadly can't see how this could help solve the problem.

\(\displaystyle \sum_{n=0}^\infty {\frac{2}{n}} \) is a divergent p-series. By comparison test, your series is divergent

[math]\sum_{n=0}^\infty {\Big(\frac{1}{n}\Big)^{a=1}} \le \sum_{n=0}^\infty {\Big(\frac{1+n^2}{1+n^3}\Big)^{a=1}}\le \sum_{n=0}^\infty {\Big(\frac{2}{n}\Big)^{a=1}} [/math]
Which p-series is convergent? Use that to conduct a comparison test with your series.

 

[ludwig259]

\(\displaystyle \sum_{n=0}^\infty {\frac{2}{n}} \) is a divergent p-series. By comparison test, your series is divergent

[math]\sum_{n=0}^\infty {\Big(\frac{1}{n}\Big)^{a=1}} \le \sum_{n=0}^\infty {\Big(\frac{1+n^2}{1+n^3}\Big)^{a=1}}\le \sum_{n=0}^\infty {\Big(\frac{2}{n}\Big)^{a=1}} [/math]
Which p-series is convergent? Use that to conduct a comparison test with your series.

Thank you, i think i get it now.
If a>1, the series on the right side is convergent, so by comparison test my series is also convergent. If a <=1, the series on the left side is divergent, meaning my series is also divergent.
I still don't think i could find these series to use comparison test with it by myself though. Would you have any advice on where to look at when using the comparison test?

 

[Deleted member 4993]

Look at response #2 & 3 together and carefully ....

 

[AvgStudent]

Thank you, i think i get it now.
If a>1, the series on the right side is convergent, so by comparison test my series is also convergent. If a <=1, the series on the left side is divergent, meaning my series is also divergent.
I still don't think i could find these series to use comparison test with it by myself though. Would you have any advice on where to look at when using the comparison test?

I think choosing which series to compare to come with experience. It gets easier as you practice (at least for me). One tip I would say about the comparison test is that it's typically used for comparing polynomials or geometric series, so I would pick the series that you're familiar with (e.g. divergent p-series in this case) and conduct the test.