<s>[INST] How to recognize a sum as a Riemann Sum Evaluate $$\frac{1}{1}+\frac{1}{2}-\frac{2}{3}+\frac{1}{4}+\frac{1}{5}-\frac{2}{6}+\frac{1}{7}+\frac{1}{8}-\frac{2}{9}+\cdots+\frac{1}{3n+1}+\frac{1}{3n+2}-\frac{2}{3n+3}+\cdots$$
answer choices:
a) $\ln 2$
b) $\ln 3$
c) $e^2$
d) $\dfrac 9 {25}$
Looking at the answer choices, I am almost certain that we need to rearrange the sum into a Riemann sum, but I am stuck on how to do so. Also are there any general techniques to recognize/rearrange summations such as these into Riemann sums? This problem is from a competition, so I would need to solve these problems quickly. Thanks!
 [/INST] Since $-\frac{2}{3n+3} = \frac{1}{3n+3} - \frac{1}{n+1}$, the general term can be written
$$a_n = \left(\frac{1}{3n+1} + \frac{1}{3n+2} + \frac{1}{3n+3}\right) - \frac{1}{n+1}$$
so the sum (up to the $n$'th term) is the difference between two harmonic numbers
$$\left(1 + \frac{1}{2}+\ldots + \frac{1}{3n+3}\right) - \left(1 + \frac{1}{2} + \ldots + \frac{1}{n+1}\right) = \sum_{k=n+2}^{3n+3}\frac{1}{k} = \frac{1}{N}\sum_{k=N+1}^{3N}\frac{1}{\frac{k}{N}}\tag{1}$$
where $N = n+1$. This is one single term $\left(\frac{1}{N} = \frac{1}{n+1}~~\text{which vanishes as $n\to\infty$}\right)$ away from being an exact Riemann sum for the integral $\int_1^3\frac{{\rm d}x}{x}$. We therefore have
$$\lim_{n\to\infty }\sum_{k=n+2}^{3n+3}\frac{1}{k} = \lim_{N\to\infty} \frac{1}{N}\sum_{k=N}^{3N}\frac{1}{\frac{k}{N}} = \int_1^3\frac{{\rm d}x}{x} = \ln(3)$$
Alternatively: if you know that the harmonic numbers satisfy $H_n \sim \gamma + \log(n)$ then applying this to $(1)$ gives us the answer directly
$$H_{3n} - H_n \sim (\gamma + \ln(3n)) - (\gamma + \ln(n)) = \ln(3)$$
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