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I am trying to prove that $\frac{1}{k+1} \le\ln(1+\frac{1}{k}) \le \frac{1}{k}$. This showed up as a known fact in a proof for $(1+\frac{1}{k})^k = e$ as $k$ goes to infinity, so I would like to prove it without using that fact.

I tried to prove the first part of the inequality by using another known inequality (one which I am comfortable proving):

$1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{k} >\ln(1+k)$

Let $a$ denote $1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{k-2}$

$a+\frac{1}{k-1}+\frac{1}{k} >\ln(1+k)$

Scaling this down 1 term:

$a + \frac{1}{k-1} >\ln(k)$

Subtracting the second inequality from the first, we get:

$\frac{1}{k} >\ln(1+\frac{1}{k})$

However, this seems to contradict the statement I'm trying to prove. Could anyone help?

Michael Rozenberg
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infinitylord
  • 4,777

4 Answers4

6

Hint:

$$\ln\left(1 + \frac1k \right) = \int_k^{k+1}\frac1t dt$$

4

The Mean Value Theorem says that for some $0\lt\xi\lt1$ $$ \log(k+1)-\log(k)=\frac1{k+\xi} $$ Therefore $$ \frac1{k+1}\lt\log\left(1+\frac1k\right)\lt\frac1k $$

robjohn
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2

Your mistake is that subtracting like inequalities doesn't tell you anything useful: for example

  • $2<4$ and $1<2$, and we do have $2-1 < 4-2$
  • $2<4$ and $1<3$, but we have $2-1 = 4-3$
  • $2<3$ and $1<3$, but we have $2-1 > 0$

Knowing $a<b$ and $c<d$ doesn't let us conclude anything about how $a-c$ and $b-d$ compare.

Subtracting opposite ones is useful, though:

Theorem: If $a<b$ and $c>d$, then $a-c < b-d$.

However, in my opinion, it's easier to remember just that adding like inequalities is fine, and convert $c>d$ to $-c<-d$ when desired.

1

A variant of Open Ball's answer:

$\ln\left(1 + \frac1 k \right) = \int_1^{1+\frac1{k}}\frac{dt} t = \int_0^{\frac1{k}}\frac{dt} {1+t} $

so $\ln\left(1 + \frac1 k \right) < \int_0^{\frac1{k}}\frac{dt} {1} =\frac1{k} $

and $\ln\left(1 + \frac1 k \right) > \int_0^{\frac1{k}}\frac{dt} {1+1/k} =\frac1{k}\frac1 {1+1/k} =\frac1{k+1} $.

marty cohen
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