How to prove that a sum is inclosed by two upper and lower bounds (values)?

Question

Show that this is true for all $n \ge 1$

$2\ln(2) - 1 \le \frac1{n} \sum_{k=1}^{n} \ln(1 + \frac{k}{n}) \le 2\ln(2) - 1 + \frac{\ln(2)}{n}$

I've no idea what to do here but here is how I've tried so far. I feel like I wanna approximate the sum by using Cauchy's integral 7,test. So I wanna rewrtie it using an integral.

$\int^{k/n} \ln (1 + x) {\rm d}x$

But the problem is I don't know what lower bound I should use. And even if I did, how do I continue? This feels like a dead end. Am I even on the right path?

Answer 1

The first half of the inequality can be solved using the AM-GM inequality.

$$\frac{1}{n}\sum_{k=1}^n\ln\left(1+\frac{k}{n}\right)$$ $$=\frac{1}{n}\ln\left(\prod_{k=1}^n\left(1+\frac{k}{n}\right)\right)$$ $$=\ln\left(\prod_{k=1}^n\left(1+\frac{k}{n}\right)\right)^{\frac{1}{n}}$$ $$=\ln(GM)$$ where $GM$ is the geometric mean of $\left(1+\frac{1}{n}\right)$, $\left(1+\frac{2}{n}\right)$, $\left(1+\frac{3}{n}\right)$, ... , $\left(1+\frac{n}{n}\right)$ $$\\$$ $$2\ln(2)-1$$ $$=\ln(2)^2-\ln(e)$$ $$=\ln(4)-\ln(e)$$ $$=\ln\left(\frac{4}{e}\right)$$ $$\approx\ln(1.4715)$$ $$\\$$ Arithmetic mean of $\left(1+\frac{1}{n}\right)$, $\left(1+\frac{2}{n}\right)$, $\left(1+\frac{3}{n}\right)$, ... , $\left(1+\frac{n}{n}\right)$ is $\left(1.5+\frac{1}{2n}\right)$ $$\\$$ $$AM \le GM$$ $$\implies \ln(AM) \le \ln(GM)$$ Since $1.4715 \le AM$, $\ln(1.4715) \le \ln(AM)$ $$\implies \ln(1.4715) \le \ln(GM)$$ Hence proving the first half of the inequality.

Answer 2

You can estimate the sum by integrals using $$\int\ln(1+x)\;dx = (1+x)\ln(1+x)-x$$

For $n=1$ the RHS is trivial and the LHS is $$\ln (1+1) \geq \int_0^1\ln(1+x)\;dx = 2\ln 2 - 1$$ Now note that for $n\geq 2$ and $1\leq k \leq n$ you have $$ \int_{\frac{k-1}n}^{\frac kn}\ln(1+x)dx\leq \frac 1n\ln\left(1+\frac kn\right) \leq \int_{\frac{k}n}^{\frac{k+1}n}\ln(1+x)dx \quad (1)$$ LHS: $$\frac 1n\sum_{k=1}^n\ln\left(1+\frac kn\right) \stackrel{(1)}{\geq} \int_0^1\ln(1+x)dx = 2\ln 2 - 1$$ RHS: $$\frac 1n\sum_{k=1}^n\ln\left(1+\frac kn\right) = \frac 1n\sum_{k=1}^{\color{blue}{n-1}}\ln\left(1+\frac kn\right) + \frac{\ln 2}n \stackrel{(1)}{\leq} \int_{\frac 1n}^1\ln(1+x)dx+\frac{\ln 2}n \leq \int_{0}^1\ln(1+x)dx+\frac{\ln 2}n = 2\ln2 - 1 + \frac{\ln 2}n$$

Answer 3

We can prove that $\,2\ln(2) - 1 \le \frac1{n} \sum_{k=1}^{n} \ln(1 + \frac{k}{n}) \le 2\ln(2) - 1 + \frac{\ln(2)}{n}$ by means of the Frullani' integral.

Denoting $\displaystyle I(n)=\frac1{n} \sum_{k=1}^{n} \ln(1 + \frac{k}{n})$ and using $\,\displaystyle \ln(1 + \frac{k}{n})=\int_0^\infty\frac{e^{-t}-e^{-(1 + \frac{k}{n})t}}{t}dt$ $$I(n)=\frac1n \sum_{k=1}^{n}\int_0^\infty\frac{e^{-t}-e^{-(1 + \frac{k}{n})t}}{t}dt=\frac1n \int_0^\infty\frac{e^{-t}}{t}\sum_{k=1}^{n}\int_0^\infty\left(1-e^{- \frac{k}{n}t}\right)dt$$ Performing summation $$I(n)=\int_0^\infty\frac{e^{-t}}{t}\Big(1-\frac{1-e^{-t}}{n(e^\frac tn-1)}\Big)dt$$ First, we notice that $\displaystyle\,e^\frac tn-1=\frac tn+\frac12\frac{t^2}{n^2}+...\geqslant\frac tn$. Hence, $$I(n)\geqslant\int_0^\infty\frac{e^{-t}}{t}\Big(1-\frac{1-e^{-t}}{t}\Big)dt=\int_0^\infty\frac{e^{-2t}+te^{-t}-e^{-t}}{t^2}dt$$ Integrating by part, $$=-\,\frac{e^{-2t}+te^{-t}-e^{-t}}{t}\bigg|_{t=0}^\infty+2\int_0^\infty\frac{e^{-t}-e^{-2t}}{t}dt-\int_0^\infty e^{-t}dt$$ Using the Frullani' integral, $$\int_0^\infty\frac{e^{-t}}{t}\Big(1-\frac{1-e^{-t}}{t}\Big)dt=2\ln2-1\tag{1}$$ Therefore, $$I(n)\geqslant2\ln 2-1\tag{2}$$ Second, we can present $I(n)$ in the form $$I(n)=\int_0^\infty\frac{e^{-t}}{t}\Big(1-\frac{(1-e^{-t})(e^{-\frac tn}-1+1)}{n(1-e^{-\frac tn})}\Big)dt$$ $$=\int_0^\infty\frac{e^{-t}}{t}\Big(1-\frac{1-e^{-t}}{n(1-e^{-\frac tn})}\Big)dt+\frac1n\int_0^\infty\frac{e^{-t}(1-e^{-t})}{t}dt$$ Using $\displaystyle 1-e^{-\frac tn}\leqslant \frac tn$ for $t\in[0;\infty)$ and (1), we get the second inequality: $$I(n)\leqslant\int_0^\infty\frac{e^{-t}}{t}\Big(1-\frac{1-e^{-t}}{t}\Big)dt+\frac1n\int_0^\infty\frac{e^{-t}-e^{-2t}}{t}dt=2\ln2-1+\frac{\ln2}{n}\tag{3}$$