Need to prove inequality $\sum\limits_{k=0}^n \frac{1}{(n+k)} \ge \frac{2}{3}$

Question

Prove that for $n \geq 1$:

$$\sum\limits_{k=0}^n \frac{1}{(n+k)} \ge \frac{2}{3}$$ I have tried math induction but that didn't work. Although I'm pretty sure that the solving can be done by induction

Answer 1

Hint: Try to show that $\dfrac{1}{n+k}+\dfrac{1}{n+(n-k)} \ge \dfrac{4}{3n}$ for all $0 \le k \le n$.

Answer 2

The given sum is an upper Riemann sum for $\int_{n}^{2n+1} \frac{dx}{x} = \log(2 + 1/n) > \log 2 > 2/3$.

Not sure if using calculus is a permitted method.

Answer 3

Denote your sum by $S_n$. Say you want to show by induction $S_n > a$ for some number $a$. That won't work directly since $S_n$ is decreasing. However, if you consider the sum from $k=1$ to $k=n$, denote it by $S'_n$, now that is increasing with $n$ since $$S'_{n+1} - S'_{n} = \frac{1}{2n+1} + \frac{1}{2n+2} - \frac{1}{n+1} >0$$ So you might just show that $S'_n >a$, or, equivalently, $S_n > a + 1/n$, for large enough $n$.

Let $a=2/3$. We calculate $S'_9 = 0.6661..$ and $S'_{10}=0.6687..$ so already $S'_{10} > 2/3$ and then, by induction, $S'_n> 2/3$ for all $n\ge 10$, or, equivalently, $S_n > 2/3 + 1/n$ for $n \ge 10$.

To complete the proof one should also check that $S_n > 2/3$ for all $1\le n \le 9$. This is easy now: $$ S_1 > S_2 > \ldots > S_{10} > S'_{10} > 2/3$$

In principle this could work for any $a < \log 2$.

Answer 4

Show the sequence is decreasing, the difference between successive terms is $$\frac{1}{2n+2}+\frac{1}{2n+1}-\frac{1}{n}$$ it is easy to prove that this is negative simply by adding together the fractions.

Now the sequence are the Riemann sums for the integral $$\int_0^1 \frac{dx}{x+1}=\ln 2$$ so the lower bound is in fact $\ln 2$

Now show that $\frac{2}{3} < \ln 2$

Answer 5

We can replace the lower bound $\frac{2}{3}$ with $\frac{15}{22}$. Since $A_1=\frac{3}{2}$ and for any $n\geq 1$ $$A_n=\sum_{k=0}^{n}\frac{1}{n+k}=H_{2n}-H_{n-1}$$ satisfies: $$ A_{n}-A_{n+1} = \frac{6n+4}{2n(2n+1)(2n+2)} \leq\frac{3}{4}\left(\frac{1}{n-1/12}-\frac{1}{n+11/12}\right)$$ we have: $$\begin{eqnarray*} A_n &=& A_1 + (A_2-A_1) + (A_3-A_2) + \ldots + (A_n-A_{n-1})\\ &\geq& \frac{3}{2}-\frac{3}{4}\sum_{j=1}^{n-1}\left(\frac{1}{j-1/12}-\frac{1}{j+11/12}\right)\geq \frac{3}{2}-\frac{3}{4}\cdot\frac{12}{11}=\frac{15}{22}>\frac{2}{3}.\end{eqnarray*}$$ To avoid non-trivial inequalities, we can just show that the sequence $\{A_n\}_{n\in\mathbb{N}^*}$ is decreasing and prove that: $$ \lim_{n\to +\infty} A_n = \int_{0}^{1}\frac{dx}{1+x} = \log 2$$ by a Riemann-sums argument. This implies $A_n\geq \log 2>\frac{9}{13}$.