Problem
Prove that $$\frac{1}{2}\ln(1+n) < \sin\left(\frac{1}{2}\right) + \sin\left(\frac{1}{4}\right) + \sin\left(\frac{1}{6}\right) + ... + \sin\left(\frac{1}{2n}\right) < \frac{1}{2}\ln(n) + \ln(2)$$
where $n$ is a positive integer.
My Process
∵ for all $x > 0$ $$\sin(x) < x$$ ∴ $$\sin\left(\frac{1}{2}\right) + \sin\left(\frac{1}{4}\right) + \sin\left(\frac{1}{6}\right) + ... + \sin\left(\frac{1}{2n}\right) < \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + ... + \frac{1}{2n}$$
∵ The integral of $1/n$ is $\ln(n)$ $$\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + ... + \frac{1}{2n} < \frac{1}{2}\ln(n)$$
But I have NO clues about how to prove the left side inequality.
Can people give me some suggestions?
With great appreciation!!
The right inequality: As already noted in the comments, your proof of the right inequality needs to be fixed: We have (see for example Bounds for the Harmonic k-th partial sum.) $$ \sum_{k=1}^n \sin\left( \frac{1}{2k}\right) < \frac 12\sum_{k=1}^n \frac 1k < \frac 12 \bigl(\ln(n) + 1 \bigr) < \frac 12 \ln(x) + \ln(2) \, . $$
The left inequality: If we can show that $$ \tag{$*$} \sin(x) > \frac 12 \ln (1+2x) $$ for $0 < x < 1$ then $$ \sum_{k=1}^n \sin\left( \frac{1}{2k}\right) > \frac 12 \sum_{k=1}^n \ln \left( \frac{k+1}{k}\right) = \frac 12 \ln (1+n) $$ is the desired estimate.
In order to prove $(*)$ we consider the function $$ \sin(x) - \frac 12 \ln (1+2x) $$ on the interval $[0, 1]$. We have $f(0) = 0$ and (using Using mean value theorem to show that $\cos (x)>1-x^2/2$) $$ f'(x) = \cos(x) - \frac{1}{1+2x} > 1 - \frac 12 x^2 - \frac{1}{1+2x} \\ = \frac{x(4-x-2x^2)}{2(1+2x)} > 0 $$ for $0 < x < 1$, so that $f$ is strictly increasing, and $(*)$ follows.