How could I show $(\frac{n+0.06}{n})^n$ is an increasing function for $\forall n\in \mathbb{N}$? I thought I'd might use induction, but that seems too hard, then I tried to take the derivative and show that that's positive $\forall$n. But I can't figure out how to do that either, I've tried induction there too.
 A: You can do it directly with the binomial formula: since \begin{align*} \Big(\frac{n+0.06}{n} \Big)^n &= 1 + \frac{0.06}{n} \binom{n}{1} + ... + \frac{0.06^n}{n^n} \binom{n}{n} \\ &= \sum_{k=0}^n \frac{0.06^k}{k!} \Big( 1 - \frac{1}{n}\Big) \cdot ... \cdot \Big( 1 - \frac{k-1}{n}\Big) \end{align*} and each $1 - \frac{m}{n+1} > 1 - \frac{m}{n}$, it follows that \begin{align*} \Big(\frac{n+1+0.06}{n+1} \Big)^{n+1} &= \sum_{k=0}^n \frac{0.06^k}{k!} \Big( 1 - \frac{1}{n+1}\Big)...\Big(1 - \frac{k-1}{n+1}\Big) \\ &\quad \quad + \underbrace{\frac{0.06^{n+1}}{(n+1)!} \Big( 1 - \frac{1}{n+1}\Big)...\Big( 1 - \frac{n}{n+1}\Big)}_{>0} \\ &> \Big(\frac{n+0.06}{n}\Big)^n.\end{align*}
A: Let's think about it via the derivative.  Let:
$$f(x) = \left(1+\frac{a}{x}\right)^x$$
When $a = 0.6$ and $x\in\mathbb N$ we have your function.
We can't take the derivative of this yet, as we can't use the power rule (the exponent is a variable), and can't use the exponential rule (the base is variable).  What we do is look at the natural log of it:
$$\ln f(x) = x\ln\left(1+\frac{a}{x}\right)$$
This is a function we can differentiate, so we get that:
$$(\ln f(x))' = \ln(1+\frac{a}{x})+x\frac{-\frac{a}{x^2}}{1+\frac{a}{x}}$$
Here, we need to be careful and recall that:
$$(\ln f(x))' = \frac{f'(x)}{f(x)}$$
This gives us that:
$$\frac{f'(x)}{f(x)} = \ln\left(1+\frac{a}{x}\right)-\frac{a}{a+x}$$
Now, we can multiply through by $f(x)$ to get that:
$$f'(x) = \left(\ln\left(\frac{x+a}{x}\right)-\frac{a}{a+x}\right)\left(1+\frac{a}{x}\right)^x$$
Now, we can rewrite:
$$\frac{a}{a+x} = 1-\frac{x}{a+x}$$
This gives us that:
$$f'(x) = \left(\ln\left(\frac{x+a}{x}\right)+\frac{x}{a+x}-1\right)\left(1+\frac{a}{x}\right)^x$$
Now, $\left(1+\frac{a}{x}\right)^x>0$ for all $x> 0$, so the sign of this will depend on the left productand. Now, we have that $\frac{x}{x+a}>\frac{1}{a}$, so as $0<a<1$, we have that $\frac{x}{x+a}>\frac{1}{a}>1$, so the left productand will be positive, so $f'(x)>0$ for $x>0$, so this result holds for $x\in\mathbb N$.
A: Let $\varepsilon = 0.06$. Now, let
$$f(x) = \left(1+\frac{\varepsilon}{x}\right)^x$$
so that your sequence is given by $f(n)$. Then
$$\log(f) = x\log\left(1+\frac{\varepsilon}{x}\right)$$
Differentiating both sides,
$$\frac{f'}{f} = \log\left(1+\frac{\varepsilon}{x}\right)-x\frac{\varepsilon/x^2}{1+\varepsilon/x} = \log\left(1+\frac{\varepsilon}{x}\right)-\frac{\varepsilon/x}{1+\varepsilon/x}$$
For $t\geq 2$ we have $\log(t) > 1/t$, and so the above is
$$>\frac{1}{1+\varepsilon/x}-\frac{\varepsilon/x}{1+\varepsilon/x} = \frac{1-\varepsilon/x}{1+\varepsilon/x}$$
Which is positive for $x> \varepsilon$. Since $f>0$, this establishes that $f$ is increasing for $x \geq 2$. You can manually check that $f(2)>f(1)$. 
