Show that
$$\sum_{n = 0}^{+\infty} \frac{\ln(n^3+1)}{n^3+1} \frac{1}{4^n} \leq \frac{\ln(2)}{6}$$
I don't know where I am wrong, but I don't get the desired result. The hint was "use the function $\frac{\ln(x^3+1)}{x^3+1}$ in the reasoning.
attempts
So I thought that if $f(x) = \frac{\ln(x^3+1)}{x^3+1}$ is limited, then I could say
$$\sum_{n = 0}^{+\infty} \frac{\ln(n^3+1)}{n^3+1}\frac{1}{4^n} \leq C \sum_{n = 0}^{+\infty} \frac{1}{4^n}$$
Which I can then sum as a Geometric Series.
So $f(x) = \frac{\ln(x^3+1)}{x^3+1}$, and by studying the derivative I got that $f(x)$ has a maximum at $x = \sqrt[3]{e-1}$ hence $$\text{max}(f(x)) = f(\sqrt[3]{e-1}) = \frac{1}{e}$$
Yet by doing so I obtain
$$\sum_{n = 0}^{+\infty} \frac{\ln(n^3+1)}{n^3+1}\frac{1}{4^n} \leq \frac{1}{e}\sum_{n = 0}^{+\infty} \frac{1}{4^n} = \frac{4}{3e}$$
But it is surely not true that $\frac{4}{3e} \leq \frac{\ln(2)}{6}$
So where I am wrong?
Thank you!
The slightly weaker bound you got was because we are actually only evaluating $f$ at positive integers. Calculus shows that $f(x)$ is decreasing for $x$ larger than the extremum you found, and comparing $f(1)$ and $f(2)$ shows us that $f(1)\geq f(n)$ for all positive integers $n$.
Then your sum can be estimated as $$\sum_{n=1}^\infty f(n)4^{-n}\leq f(1)\sum_{n=1}^\infty 4^{-n}=\frac{\log 2}{6}. $$