# Example of a convergent series for which integral test fails?

Is there example of a convergent series for which integral test fails or can not be applied? Just wondering if integral test is the silver bullet of convergence tests, or are there any series that any of the other tests can determine convergence but integral test fails?

Of course by convergence we mean classical convergence, not Cesàro summable series that do not have a classical sum.

• Integral test never fail (if applicable). It's a if and only if statement. – user99914 May 26 '15 at 11:41
• @John : so why use the other tests when integral test is enough? – Arjang May 26 '15 at 11:42
• Because sometimes it is hard to calculate the integral. – user99914 May 26 '15 at 11:43
• Because it's not always applicable. It only works for monotone decreasing function. Besides the integral is not always easy to compute. – Najib Idrissi May 26 '15 at 11:43
• @Arjang because sometimes it becomes too complicated to evaluate the integral ,other tests often come handy in those situations – Taylor Ted May 26 '15 at 11:43

No, there isn't. As long as you can apply the integral test, then it has to work: if $f : [0, \infty) \to [0, \infty)$ is decreasing, then the series $\sum f(n)$ converges iff the improper integral $\int_0^\infty f(t) dt$ converges. This isn't, say, like the root test or the ratio test that can be inconclusive, even when you can apply it (when $\lim a_{n+1}/a_n = 1$ or $\limsup |a_n|^{1/n} = 1$).
But there are two caveats here. First, this only works for nonnegative, decreasing functions $f$. If your function takes negative values, or if it is not decreasing, then you cannot apply the integral test. In that sense, it is not a "silver bullet". Another caveat is that you need to compute an integral to apply the test (or at least determine if it's convergent), and this is not always easy or even feasible. Determining whether $\int_0^\infty f(t) dt$ converges can be just as hard as determining if $\sum f(n)$ converges.