You know that \(\displaystyle \sum_{n=0}^\infty {\frac{1}{n}} \) is called a harmonic series and it it divergent.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.
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View attachment 30747
True, I sadly can't see how this could help solve the problem.You know that \(\displaystyle \sum_{n=0}^\infty {\frac{1}{n}} \) is called a harmonic series and it it divergent.
\(\displaystyle \sum_{n=0}^\infty {\frac{2}{n}} \) is a divergent p-series. By comparison test, your series is divergentTrue, I sadly can't see how this could help solve the problem.
Thank you, i think i get it now.\(\displaystyle \sum_{n=0}^\infty {\frac{2}{n}} \) 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.
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.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?