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!
 A: You actually don't have to recognize anything, you may leave the computations do that for you.
We have:
$$\begin{eqnarray*} \sum_{k\geq 0}\left(\frac{1}{3k+1}+\frac{1}{3k+2}-\frac{2}{3k+3}\right)&=&\int_{0}^{1}(1+x-2x^2)\sum_{k\geq 0}x^{3k}\,dx\\ &=& \int_{0}^{1}\frac{1+x-2x^2}{1-x^3}\,dx\tag{1} \end{eqnarray*}$$
hence the original series is converted into:
$$ \int_{0}^{1}\frac{1+2x}{1+x+x^2}\,dx =\left[\log(1+x+x^2)\right]_{0}^{1}=\color{red}{\log 3}.\tag{2}$$

As a time-saving alternative, you may notice that by summing just the first six terms you get a number that is close to $1$, and $(B)$ is the only option close to $1$.
A: Since each "chunk" of three consecutive terms consists of
$$\frac{1}{3n+1}+\frac{1}{3n+2}-\frac{2}{3n+3}$$
Then you should combine them into a single term:
$$\frac{(3n+1)(3n+3)+(3n+2)(3n+3)-2(3n+1)(3n+2)}{(3n+1)(3n+2)(3n+3)}$$
which condenses down to
$$\frac{9n+5}{(3n+1)(3n+2)(3n+3)}$$
And so your Riemann Sum should be
$$\sum_{n=1}^\infty \frac{9n+5}{(3n+1)(3n+2)(3n+3)}$$
Now, if you need a hint on how to actually evaluate it once you have it as a Riemann Sum, here is what you should do. Let $S(a)$ be a function defined as
$$S(a)=\sum_{n=1}^\infty \frac{(9n+5)a^{3n+3}}{(3n+1)(3n+2)(3n+3)}$$
Then, as you can see, when you find $S'''(a)$, you get
$$S'(a)=\sum_{n=1}^\infty \frac{(9n+5)a^{3n+2}}{(3n+1)(3n+2)}$$
$$S''(a)=\sum_{n=1}^\infty \frac{(9n+5)a^{3n+1}}{(3n+1)}$$
$$S'''(a)=\sum_{n=1}^\infty (9n+5)a^{3n+1}$$
This is easier to evaluate. Once you have a formula for this, you need only integrate $S'''(a)$ three times to find $S(a)$ in terms of $a$. Then notice that your sum is equal to $S(1)$.
However, since you mention that these are contest problems that would need to be solved quickly, here is what my strategy would be: you can rule out $A$ and $C$ immediately, since $A$ is the value of the alternating harmonic series (which this is obviously not), and since there are no factorials in the sum, suggesting that $e$ is not involved in the final value.
Now you must decide between $B$ and $D$. Notice that our $S'''(a)$ would turn into a geometric series should we decide to evaluate it. This means that some part of $S'''(a)$ should contain the expression
$$\frac{1}{1-a}$$
which suggests that the integrals of $S'''(a)$ would contain logarithms. Thus the answer should (I'm guessing) be $\ln 3$.
Am I right?
A: 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
$$H_{3n} - H_n \sim (\gamma + \ln(3n)) - (\gamma + \ln(n)) = \ln(3)$$
