Prove that $\displaystyle \sum_{k=1}^n \bigg(\dfrac{1}{k}+\dfrac{2}{k+n}\bigg ) \leq \ln(2n) + 2 -\ln(2)$

Question

Prove that $$\displaystyle \sum_{k=1}^n \bigg(\dfrac{1}{k}+\dfrac{2}{k+n}\bigg ) \leq \ln(2n) + 2 -\ln(2).$$

I was thinking of using mathematical induction for this. That is,

We prove by induction on $n$. The case $n=1$ holds trivially since $2 \leq 2$. Now assume the result holds for some $m$. Then by assumption we know that $$\displaystyle \sum_{k=1}^{m+1} \bigg(\dfrac{1}{k}+\dfrac{2}{k+m}\bigg ) \leq \ln(2m) + 2 -\ln(2)+\dfrac{1}{m+1}+\dfrac{2}{2m+1}. $$

We must relate this somehow to $\ln(2(m+1)) + 2 -\ln(2)$.

Answer 1

First note, that we can rewrite the RHS of your inequality as $\ln(n) +2$.

We proceed by induction. We compute

$$ \sum_{k=1}^{n+1} \left( \frac{1}{k} + \frac{2}{k+n+1}\right) = \frac{1}{n+1} + \frac{2}{2n+2} + \sum_{k=1}^n \frac{1}{k} + \sum_{k=1}^{n} \frac{2}{k+n+1}$$

shifting the index yields

$$ =\frac{2}{n+1} + \sum_{k=1}^n \frac{1}{k} + \sum_{k=2}^{n+1} \frac{2}{k+n} = \sum_{k=1}^n \frac{1}{k} + \sum_{k=1}^{n+1} \frac{2}{k+n} = \sum_{k=1}^n\left( \frac{1}{k} + \frac{2}{k+n}\right) + \frac{2}{2n+1}$$

applying the induction hypothesis gives

$$ \leq \ln(n) + 2 + \frac{2}{2n+1} = \ln(n+1) +2 + \left( \frac{2}{2n+1} + \ln(n) - \ln(n+1) \right).$$

Hence, we are left to show that

$$ \frac{2}{2n+1} - \ln\left(1 + \frac{1}{n}\right) = \frac{2}{2n+1} + \ln(n) - \ln(n+1) \leq 0. $$

This is already done in a previous answer.

Answer 2

If you use induction, then what you need to prove in the end will be simplified to the following: $$\dfrac{2}{2n+1}\leq\log\big(1+\dfrac{1}{n}\big)$$. I suggest you make sure to get this when you do the induction step.

In general, following is true: $$\log(1+x)\geq\dfrac{2x}{x+2}, \,x\geq 0$$. In fact, let $f(x) = \log(1+x) - \dfrac{2x}{x+2}$, then $f'(x) = \dfrac{1}{1+x} - \dfrac{4}{(x+2)^2} = \dfrac{x^2}{(1+x)(x+2)^2}\geq 0$. Hence, $f(x)\geq f(0) = 0 - 0 = 0.$ Now put $x = \frac{1}{n}$, and you get the desired result.

Note that this is slightly stronger than the well-known bound $\log(1+\dfrac{1}{n})\geq\dfrac{1}{n+1}$.

Answer 3

If you want to do a proof by induction. you have covered the base case.

Inductive hypothesis:

$\sum_\limits{k=1}^n \bigg(\dfrac{1}{k}+\dfrac{2}{k+n}\bigg ) \leq \ln(2n) + 2 -\ln(2)$.

We need to show that:

$\sum_\limits{k=1}^{n+1} \bigg(\dfrac{1}{k}+\dfrac{2}{k+n+1}\bigg ) \leq \ln(2n) + 2 -\ln(2)$

thoughts:

$\ln(2n) + 2 -\ln(2) = \ln (n) + 2$

$\sum_\limits{k=1}^{n+1} \bigg(\dfrac{1}{k}+\dfrac{2}{k+n+1}\bigg ) = \sum_\limits{k=1}^{n} \bigg(\dfrac{1}{k}+\dfrac{2}{k+n}\bigg ) + \dfrac{2}{2n+1}$

$\sum_\limits{k=1}^{n} \bigg(\dfrac{1}{k}+\dfrac{2}{k+n}\bigg )\leq \ln n + 2$ by the inductive hypothesis.

So, if $\dfrac{2}{2n+1}-\ln (n) + 2 \leq \ln(n+1) + 2$, we are done.

$\dfrac{2}{2n+1} < \ln(1+\frac{1}{n})$

$\sum_\limits{k=1}^{n+1} \bigg(\dfrac{1}{k}+\dfrac{2}{k+n+1}\bigg ) \leq \ln(2n) + 2 -\ln(2)$

QED

Answer 4

A non-induction version, just for diversification. Let's note $$S_n=\sum_{k=1}^{n} \frac{1}{k}$$ which is $$S_n=\ln(n)+\gamma +\varepsilon_n$$ Now $$\sum_{k=1}^{n}\left ( \frac{1}{k} + \frac{2}{k+n} \right )= \sum_{k=1}^{n} \frac{1}{k} + \sum_{k=1}^{n} \frac{2}{k+n} $$ $$ = 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n} + 2\left ( \frac{1}{n+1} + \frac{1}{n+2} + ... + \frac{1}{2n} \right ) $$ $$=S_n+2S_{2n}-2S_n=\ln(n)+\gamma +\varepsilon_n + 2\ln(2n)+2\gamma +2\varepsilon_{2n} - 2\ln(n)-2\gamma -2\varepsilon_n$$ $$=\ln(n)+\gamma + 2\ln2 + 2\varepsilon_{2n} -\varepsilon_n < $$ $$<\ln(n)+1.96352 + 2\varepsilon_{2n} -\varepsilon_n$$ And $$2\varepsilon_{2n} -\varepsilon_n \sim \frac{2}{4n} - \frac{1}{2n}$$ i.e. first $n=1..4$ can be verified "manually".