My question isn't about a specific mathematical problem. It's about the comparison method in general. I've been studying some real analysis and I specially struggle with the convergence test problems, and have realised that most of them can be solved easily with comparison test. I totally understand the method and how it works, but I have this questions:

  • Is there any "typical form" of series $\sum a_n$ that I can recognize and say: "This one is about comparison test" ? (For example, for the series $\sum (a_n)^n$ I intuitively go for the root test, so I want to find a pattern like that but for comparison test).
  • Is there any "rule" or method I can follow to easily find the right sequences to compare with? (I've seen answers in my textbook that use comparison test with sequences that seem totally impossible to guess, unless I'm missing something about how to find that "good" sequences).

I'm not pretty sure if this will make any sense, but I struggle with this method of convergence testing and maybe there's something I'm ignoring that makes it easier. Any help will be very appreciated.

  • 3
    $\begingroup$ The trick is to understand the asymptotic behavior of the general term, i.e. a simpler expression which is similar in the long run. Like $\frac n{n^2+1}\sim\frac1n$. $\endgroup$ – user65203 Jul 5 '20 at 16:29

If your series is of the form$$\sum_{n=1}^\infty\frac{a_n}{b_n},\tag1$$and each $a_n$ and each $b_n$ is the sum of several things, try to see which of those things grow faster; then compare $(1)$ and the series $\sum_{n_1}^\infty\frac{\alpha_n}{\beta_n}$, where $\alpha_n$ and $\beta_n$ are those things that grow faster.

For instance, consider the series$$\sum_{n=1}^\infty\frac{n^3+3n^2+\sqrt n}{2n^5+10n^3+\log(n)}.\tag2$$The part of the numerator which grows faster is $n^3$, and the part of the denominator which grows faster is $2n^5$. So, compare the series $(2)$ with the series $\sum_{n=1}^\infty\frac1{2n^2}\left(\text{i.e.},\sum_{n=1}^\infty\frac{n^3}{2n^5}\right)$, which converges. Since you have$$\lim_{n\to\infty}\frac{\frac{n^3+3n^2+\sqrt n}{2n^5+10n^3+\log(n)}}{\frac1{2n^2}}=\lim_{n\to\infty}\frac{n^5+3n^4+\sqrt n\,n^2}{2n^5+10n^3+\log(n)}=1,$$your series converges too.

  • $\begingroup$ I get what your're saying. Thank you very much for your help! $\endgroup$ – Alejandro Bergasa Alonso Jul 5 '20 at 16:19

There is no secret recipe; however some guidelines that I use include these:

1.If you see $n!$ it is a safe bet to use the ratio test ($\lim_{n\to\infty}\frac{a_{n+1}}{a_n}$)

2.Use a healthy suspicion regarding the series $a_n$. Does it even converge to zero? Usually very easy to check.

3.Using the integral or sum test with a simple function, like $\sum{1/n^{2}}$

4.Some series are just "pre-cooked" to be convergent via a theorem such as Dirichlet.

Those are from the top of my head; however, it just takes time and patience to learn to recognize what works and what doesn't. Sorry if it isn't very helpful, but I don't think there's any better insight than practice.

  • $\begingroup$ That kind of patterns or guidelines are exactly what I'm trying to find. Thanks for your answer! $\endgroup$ – Alejandro Bergasa Alonso Jul 5 '20 at 16:17

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