Prove $\frac{1}{n+1}\le \log\left(1+\frac{1}{n}\right)\le \frac{1}{n} ,\forall n\ge1$ if $\log x= \int_{1}^{x}\frac{dt}{t},x>0$

Question

I know it is very simple but something is going amiss.

We can see here that $\log(1+\dfrac{1}{n})=\displaystyle \int_{n}^{n+1}\dfrac{dt}{t}$

if $t \in[n,n+1]$, $\dfrac{1}{n+1}\le\dfrac{1}{t}\le\dfrac{1}{n}$

I want to show that $\dfrac{1}{n+1}\le\displaystyle \int_{n}^{n+1}\dfrac{dt}{t}\le\dfrac{1}{n}$.

Can I straight infer that? If so what is the logic behind it? I was thinking of using $\displaystyle \Sigma_{i=1}^{n}$ and sandwiching $\displaystyle \int_{n}^{n+1}\dfrac{dt}{t}$ between. But confused !

$\frac{1}{n+1} \le \frac 1t \le \frac 1n$ for all $t$ in the interval ... — Martin R, Apr 28 '18 at 18:54
Exactly if $t \in [n,n+1]$, I thought it too that way only. Am I right ? — Saradamani, Apr 28 '18 at 18:55
If you have to ask what is the logic behind it, you cannot straight infer it. — Hagen von Eitzen, Apr 28 '18 at 18:57
@HagenvonEitzen See I changed my edits there, I assumed it without putting it there. It seemed to me quite obvious. That's why I asked. And I said "Can I infer it?" — Saradamani, Apr 28 '18 at 18:58
like this one:https://math.stackexchange.com/questions/1887583/show-that-for-all-n-1-frac1n-1-log1-frac1n-frac1n. — StubbornAtom, Apr 28 '18 at 19:22
I have changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance. — GNUSupporter 8964民主女神地下教會, Apr 28 '18 at 19:45

score 2 · Answer 1 · answered Apr 28 '18 at 18:56

Yes, because $\frac{1}{n+1} \leq \frac{1}{t} \leq \frac{1}{n}$ for $t \in [n, n+1]$, so $$\int_{n}^{n+1} \frac{1}{n+1} \, dt \leq \int_{n}^{n+1} \frac{1}{t} \, dt \leq \int_{n}^{n+1} \frac{1}{n} \, dt$$ then your inequality follows.

score 1 · Accepted Answer · answered Apr 28 '18 at 19:02

1

$$n\le t\le n+1$$

$$\frac {1}{n+1} \le \frac {1}{t}\le \frac {1}{n}$$

$$ \int _n ^{n+1}\frac {dt}{n+1} \le \int _n ^{n+1}\frac {dt}{t}\le \int _n ^{n+1}\frac {dt}{n}$$ $$\dfrac{1}{n+1}\le\displaystyle \int_{n}^{n+1}\dfrac{dt}{t}\le\dfrac{1}{n}$$

answered Apr 28 '18 at 19:02

Mohammad Riazi-Kermani

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Prove $\frac{1}{n+1}\le \log\left(1+\frac{1}{n}\right)\le \frac{1}{n} ,\forall n\ge1$ if $\log x= \int_{1}^{x}\frac{dt}{t},x>0$

2 Answers2