Find the Following Sum!

Expressed with sigma notation, the expression is equivalent to $\displaystyle\lim_{N\to\infty}{\displaystyle\sum_{i=0}^{N}{\frac{1}{2N+i}}}$ .

An easy way to solve this would be to use integrals. To do that, we first need to manipulate this (the reason should hopefully be clear soon): $\displaystyle\lim_{N\to\infty}{\displaystyle\sum_{i=0}^{N}{\frac{1}{N}\frac{1}{2+\frac{i}{N}}}}$ .

Define $x=\frac{i}{N}$ . For sigma notation, each increment is by $1$ . Thus, $dx=\frac{i+1}{N}-\frac{i}{N}=\frac{1}{N}$ .

In addition, the bounds of the sigma are from $i=0$ to $N$ . Thus, the bounds for $x$ , which is defined as $x=\frac{i}{N}$ , are from $x=0$ to $1$ .

In integral notation, the original sum is then $\displaystyle\int_{0}^{1}\frac{dx}{2+x}$ .

(If this does not make sense, remember that $\displaystyle\int_{a}^{b}f(x)\ dx=(b-a)\displaystyle\lim_{N\to\infty}\displaystyle\sum_{i=0}^{N}\frac{f(a+\frac{(b-a)i}{N})}{N}$ and work from there.)

Evaluate the integral: $\displaystyle\int_{0}^{1}\frac{dx}{2+x}=[\ln(2+x)]_{0}^{1}=\ln(3)-\ln(2)=\ln(\frac{3}{2})$ .

The final answer is thus $3+2=5$ .

Relevant wiki: Riemann Sums

$\begin{aligned} L & = \lim_{N \to \infty} \left(\frac 1{2N} + \frac 1{2N+1} + \frac 1{2N+2} + \cdots + \frac 1{3N}\right) \\ & = \lim_{N \to \infty} \sum_{k=0}^N \frac 1{2N + k} & \small \color{#3D99F6} \text{Dividing up and down by }N \\ & = \lim_{N \to \infty} \sum_{k=0}^N \frac 1N \times \frac 1{2 + \frac kN} & \small \color{#3D99F6} \text{By Riemann sums: } \lim_{n \to \infty} \frac 1n \sum_{k=a}^n f \left( \frac kn \right) = \int_0^1 f(x) \ dx \\ & = \int_0^1 \frac 1{2+x} dx = \ln (2+x) \bigg|_0^1 = \ln \frac 32 \end{aligned}$

$\implies a + b = 3 + 2 = \boxed{5}$