If $$n\ge 2,$$ then the value of $$1+\frac 12+\frac 13+...+\frac1n$$ is​

Correct option is C

Solution:

​$$H(n) = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}, \quad \text{where } n \geq 2\\[10pt]H(n) > \frac{2n}{n + 1}\\[10pt]\begin{aligned}&\text{The harmonic series grows slowly but satisfies the inequality:} \\&H(n) > \frac{2n}{n + 1} \text{ for all } n \geq 2 \\&\text{Example: For } n = 2, \\&H(2) = 1 + \frac{1}{2} = \frac{3}{2} \\&\frac{2n}{n + 1} = \frac{4}{3} \\&\frac{3}{2} > \frac{4}{3}, \text{ so the inequality holds.} \\&\text{This pattern continues for larger values of } n.\end{aligned}$$​​

​​Final Answer:
​$$\boxed{\text{(c) } > \frac{2n}{n + 1}}$$​​