Series $A = \sum_{n=1}^\infty\frac{1}{\ln(n+1/n)}$ diverges by the comparison test (wolfram). I want to compare $\sum_{n=1}^\infty\frac{\sin^4n}{\ln(n+1/n)}$ with series $A$.
How can I prove that series $A$ diverges? Or maybe there are other methods?
Note that $$ \sin^4 n = (\sin^2 n)^2 = \left( \frac{1 - \cos 2n }2 \right)^2 = \frac{1 - 2\cos 2n + \cos^2 2n}4 $$ $$ = \frac{1-2\cos 2n}4 + \frac{1+2\cos 4n}8 = \frac38-\frac{\cos 2n}2+\frac{\cos 4n}4. $$ We split the sum into three parts: $$ \sum_{n=1}^{\infty} \frac{\sin^4 n }{\ln(n+\frac1n)} = \sum_{n=1}^{\infty} \frac38 \frac1{\ln(n+\frac1n)} - \sum_{n=1}^{\infty} \frac{\cos 2n}{2\ln(n+\frac1n)} + \sum_{n=1}^{\infty} \frac{\cos 4n}{4\ln(n+\frac1n)}=\Sigma_1 - \Sigma_2 + \Sigma_3. $$ It is easy to see that $\Sigma_1$ diverges by a comparison with $\sum 1/n$.
For the remaining sums, consider the bounded partial sums $$ S_{1,N}=\sum_{n\leq N} \cos 2n, \ \ S_{2,N} = \sum_{n\leq N} \cos 4n. $$ Then by Dirichlet's Test, the two sums $\Sigma_2$ and $\Sigma_3$ are convergent.
Thus, by combining three sums, we see that the series in question is divergent.
First, note that there exists $\delta$ such that $|\sin^4(x)|<\delta$ implies $|\sin^4(x+1)|>\delta$. To see this, draw a circle. The angles $x$ for which $|\sin(x)|<\delta$ correspond so arcs around $(1,0)$ (at angle $0$) and $(-1,0)$ (at angle $\pi$), so if these arcs are small enough, which corresponds to $\delta$ being small, then adding $1$ to some angle inside these arcs will go to a new point outside these arcs.
In fact, any $\delta<\sin((\pi-1)/2))^4$ works. I will let you work out the details. Anyway, fix some $\delta$ with this property.
By the choice of $\delta$, for each $n$ we can pick a number $k_n\in\left\{n,n+1\right\}$ with $\sin^4(n_k)>\delta$. Then $$\sum_{n=1}^\infty\frac{\sin^4(n)}{\ln(n+1/n)}\geq\delta\sum_{n=1}^\infty\frac{1}{\ln(k_n+1/k_n)}$$ so we simply need to show that the RHS diverges.
Lemma: Let $(a_n)$ be a decreasing sequence of positive numbers. For each $n$, pick a number $k_n\in\{n,n+1\}$. Then $\sum_n a_n$ converges iff $\sum_n a_{k_n}$ converges.
To finish, apply the lemma above with $a_n=\frac{1}{\ln(n+1/n)}$.
Proof: Suppose $\sum_n a_{k_n}$ converges. For each $n$, let $p_n$ be the only element of $\{n,n+1\}\setminus\{k_n\}$. Since $a_n$ is decreasing, $k_n>p_{n+1}$. Therefore, $$\sum_n a_n=\sum_n a_{p_n}+a_{k_n}\leq a_{p_1}+2\sum_n a_{k_n}<\infty.$$ The other implication is trivial. QED
Note that $\sin^4(t) + \sin^4(t+1)$ is continuous and nonnegative, is never $0$ and is periodic. It follows that there is a constant $c> 0$ such that $\sin^4(t) + \sin^4(t+1) > c$ for all $t.$ Also note that $\sin^4(n)/\ln (n+1/n) \ge \sin^4(n)/\ln (2n).$ So it's enough to show the series $\sum \sin^4(n)/\ln (2n) = \infty.$ Here observe that
$$ \frac{\sin^4(2n)}{\ln (2(2n))} + \frac{\sin^4(2n+1)}{\ln (2(2n+1))} \ge \frac{\sin^4(2n) + \sin^4(2n+1)}{\ln (4n+2)} \ge \frac{c}{\ln (4n+2)}.$$
Since $\sum_{n=1}^{\infty} c/\ln (4n+2) = \infty,$ the original series diverges too.
