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There are several convergence tests (Ratio, Root, Direct comparison, Integral, etc). When one test is inconclusive, another more powerful one (or a combination of tests) is used.

For example, the convergence of the series $\sum_{n=1}^{\infty}\frac{\ln n}{n^2}$ (Bertrand series) is inconclusive by Ratio and Root tests, but is proved by the Comparison test ($\ln n\le \sqrt{n}$). Or it can be tested by Integral test (by integration by parts). Or it can be tested by Ermakoff's test (see here or here (in Russian, page $257$, Q.$2624$)):

$$\lim_\limits{x\to\infty}\frac{e^xf(e^x)}{f(x)}=\lim_\limits{x\to\infty}\frac{e^x\cdot \frac{x}{e^{2x}}}{\frac{\ln x}{x^2}}=\lim_\limits{x\to\infty}\frac{x^3}{e^x\ln x}=0<1.$$ Now my question is: can a series be made so that the Comparison and Integral tests are very difficult and the Ermakoff's test is advantageous?

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  • $\begingroup$ it's quite hard to find something difficult for the comparison test, considering most Ermakoff problems involve logarithms, which are relatively easy to compare. however, $$ \sum \frac{\ln n}{n \ln \ln n} $$ can be solved via Ermakoff, the integral test is useless because of the exponential integral. $\endgroup$ – Dando18 Jun 20 '17 at 13:59
  • $\begingroup$ The Comparison Test is not a by a specific formula. If $|a_n|\leq |b_n|$ for all $n$ and you already know that $ \sum_n|b_n|$ converges then you know that $\sum_na_n$ converges. So the degree of difficulty of the Comparison Test depends on what catalog of known absolutely convergent series is available to you. $\endgroup$ – DanielWainfleet Jun 21 '17 at 6:23

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