Show this function is bounded by 1/n Let  $b>0$ and $f(n) = (\frac{3+2n}{1+2n})^b-1$. I need to show $f(n) \in O(\frac{1}{n})$. Is this even true?
 A: it is true and here is why: let $N$ be the least integer greater than $b$. It is $$\bigg{(}\frac{3+2n}{1+2n}\bigg{)}^b-1=(1+\frac{2}{1+2n})^b-1\leq(1+\frac{2}{1+2n})^N-1<\sum_{k=1}^N{N\choose k}\frac{1}{n^k}\leq$$ $$\leq M\cdot (\frac{1}{n}+\frac{1}{n^2}+\dots+\frac{1}{n^N})\leq M\cdot N\cdot\frac{1}{n}$$ where $M$ is the maximum of the binomial coefficients appearing in the sum.
Note that the first inequality is true, since $t\mapsto a^t$ is increasing if $a>1$.
A: You can proceed as follows using the MVT:


*

*Write $f(n) = (\frac{\frac{3}{n}+2}{\frac{1}{n}+2})^b-1$ and

*consider $g(x) = \left(\frac{2+3x}{2+x}\right)^b = \left(1+\frac{2x}{2+x}\right)^b$ for $x \to 0^+$

*Note that $g(0) = 1$ and $g(1/n) - g(0) = f(n)$

*Furthermore, you have $g'(x) = b\left(1+\frac{2x}{2+x}\right)^{b-1}\frac{4}{(2+x)^2} \leq C$ for $x \in [0,1]$, since $g'$ is continuous on $[0,1]$
According to MVT you have for $n>1$ with $x_n = \frac{1}{n} \in (0,1)$
$$f(n) = g(\frac{1}{n}) - g(0) \stackrel{\xi_n \in (0,\frac{1}{n})}{=} g'(\xi_n )\frac{1}{n} \leq C \frac{1}{n}$$
A: $$y=\left(\frac{2 n+3}{2 n+1}\right)^b\implies \log(y)=b \log\left(\frac{2 n+3}{2 n+1}\right)=b \log\left(1+\frac{2}{2 n+1}\right)$$ Using Taylor expansions
$$\log(y)=b\left(\frac{1}{n}-\frac{1}{n^2}+O\left(\frac{1}{n^3}\right)\right)$$ Now, Taylor again
$$y=e^{\log(y)}=1+\frac{b}{n}+\frac{(b-2) b}{2 n^2}+O\left(\frac{1}{n^3}\right)$$
