Problem: Determine whether the following series converges. Solution: For positive integers k, the ratio 1/k is positive and at most 1, so the sine and a general term of the series is positive as well. Thus we have a series with non-negative terms and we can start choosing the right test.

There is no obvious angle to this series. Integrating does not seem easy, taking the k-th root of sine does not look good, working out ratio of complicated sines either, and there seem to be no parts to ignore for comparison. So where do we start? We usually have good results with the Root test. As expected, the sine part leads to trouble, namely to an indeterminate power. We can work it out for instance like this: So after all that hard work we got the indecisive case ϱ = 1. Since the constant λ for the Ratio test is bound to be also 1, there is no point in doing it, but out of curiosity we will have a look. This actually seemed a bit easier. Given that these two tests did not help, what else can we try? The idea of integrating the given expression is not too appealing, but we are slowly getting desperate and after all, sine and powers are not all that bad. Can we use the Integral test? Yes, since the function f (x) = sin(1/x)/x is both positive and decreasing on [1,∞). So we pass to integral and if you try to figure it out, sooner or later you start having the feeling that methods of integration covered here are not good enough to handle it. In fact, the problem is not with the methods but with this function. As a continuous function it has some antiderivative, but it is known that there is no way to write this antiderivative using a formula composed of elementary functions and operations, so there is no way we can integrate. So much for Integral test, but see Remark below.

What is left is comparison tests. No part of the given expression can be ignored for large k, but perhaps we can do with an estimate, the sine suggests a natural upper bound. For k a natural number we have that 1/k lies between 0 and 1, so sine of it is positive and we therefore have We have a divergent series on the right (it is the famous harmonic series or use the p-test) and the given series is smaller, which is exactly the situation when comparison does not help.

Is there something else we can do? There is one test we did not try, the limit comparison. It does not seem very suitable, but we are getting desperate. What do we know about the terms ak when k gets really large? Then 1/k gets really small and we know that for small y the values of sin(y) are basically equal to y. Thus we may guess that Of course, the Limit comparison test requires that we justify our guess, especially if it is not exactly traditional. So our guess was correct and we can proceed with the comparison. Since the test series on the right converges (see the p-test), so does the series on the left.

Conclusion: The given series converges.

We can get the same conclusion by the plain Comparison test, but we have to tighten up on the estimate we tried before, it was too generous. We know that for any positive y we have sin(y) < y. Thus we can do and the conclusion follows from the fact that the larger test series converges.

Remark. Here we look closer at the integral from the Integral test. First, note that It is well-known that the function on the right does not have an antiderivative that can be expressed using elementary functions and operations. While this is the end of "classical" integration, we may try to approach this problem via power series. Using the standard expansion of sine we get However, we can switch the integral and summation only on sets where the series converges uniformly, see Series of functions in Theory - Series of functions. Fortunately for us, here we have uniform convergence on the interval (0,1], so we can use the above result on any interval of the form [A,1] for A positive, and that is exactly what we need to evaluate the relevant improper integral. The resulting series is convergent, which is easy to show using the Alternating series test. So also the Integral test confirms that the given series converges, but it was so much trouble that the comparison approach is definitely better.