Question: How can I prove that there is $c>0$ such that starting from a specific $n$ the following is true:
$$
\log\_2(n+1) - \log\_3(2n+1) \le c
$$

Answer: 
**HINT**
$$
\log\_2(n+1) -\log\_3(2n+1) \le \log\_2(n+1)
$$
so suffices to prove that $\log\_2(n+1) \le cn$ for some $c \in \mathbb{R}^+$, which is equivalent to proving
$$
n+1 = 2^{\log\_2(n+1)} \le 2^{cn} = \left(2^c\right)^n = k^n
$$
for $k=2^c \in \mathbb{R}^+$.

Can you finish?

---

**UPDATE**

Now that you updated the question, it became a more interesting problem. One way is to consider the function
$$
f(x)
 = \log\_2(x+1) - \log\_3(2x+1)
 = \frac{\ln(x+1)}{\ln 2} - \frac{\ln(2x+1)}{\ln 3}
$$
over $x \ge 0$ and prove that the function is bounded by showing that it achieves a maximum or reaches a horizontal asymptote.

Achieving maximum (if true) can be shown by solving $f'(x)=0$ and computing $f''(x)$ to establish concavity with the 2nd derivative test.

Horizontal asymptotes are also part of standard calculus class.

Can you finish?

---

**UPDATE 2**

It appears this claim is not true. Note that as $x \to \infty$, we have
$$
\begin{split}
f(x)
 &\approx \log\_2(x) - \log\_3(2x) \\
 &= \log\_2(x) - \log\_3(x) - \log\_3 2 \\
 &= \frac{\ln x}{\ln 2} - \frac{\ln x}{\ln 3} - \log\_3 2 \\
 &= \ln x \left(\frac{1}{\ln 2} - \frac{1}{\ln 3}\right) - \log\_3 2 \\
 &= a\ln(x) - b,
\end{split}
$$
where $a = \frac{1}{\ln 2} - \frac{1}{\ln 3} > 0$, so $f(x) \to \infty$ as $x \to \infty$. That means the difference between your logarithmic factors diverges with increasing $n$...
$\log\_2(n+1)-\log\_3(2n+1)=\dfrac{\log\_3(n+1)}{\log\_32}-\log\_3(2n+1)=$

$=\log\_23\cdot\log\_3(n+1)-\log\_3(2n+1)=$

$=\left[\log\_2\left(\dfrac32\right)+1\right]\cdot\log\_3(n+1)-\log\_3(2n+1)=$

$=\log\_2\left(\dfrac32\right)\cdot\log\_3(n+1)+\log\_3(n+1)-\log\_3(2n+1)=$

$=\log\_2\left(\dfrac32\right)\cdot\log\_3(n+1)+\log\_3\left(\dfrac{n+1}{2n+1}\right)>$

$>\log\_2\left(\dfrac32\right)\cdot\log\_3(n+1)+\log\_3\left(\dfrac12\right)\to+\infty$

as $\;n\to+\infty\;.$

Hence, there does not exist any $\;c>0\;$ such that

$\log\_2(n+1)-\log\_3(2n+1)\le c\;$ eventually.