# How might one prove the following inequality?

Let $r$ be a natural number. I wish to prove that $x + \dfrac{x^2}{2} + \dfrac{x^3}{3} + \cdots + \dfrac{x^r}{r} \leq x^{r+1} + \log(r+1)$ for all $x>0$. Some friends and I have tried using various calculus techniques but none of them seem to work. A solution or hint would be appreciated.

• Since $r$ is a natural number this problem lends itself to a proof by induction Sep 22, 2015 at 20:57
• Yes, we tried this but the inductive step is not obvious (at least not to us). Do you have a solution in mind or were you simply suggesting a general technique? Sep 22, 2015 at 21:13
• Just to be sure: $\log$ is $\log_{10}$ or $\log_e=\ln$?
– f10w
Sep 23, 2015 at 15:32
• By $\log$, I mean the natural logarithm $\log_e$. Sep 23, 2015 at 20:10
• I can do it for $x \ge 1$ but I'm having trouble with $x<1.$ Sep 25, 2015 at 1:10

## 1 Answer

Let $f(x)=x^{r+1}+\log(r+1)-\sum_{i=1}^r{x^i/i}$. We need to show that $\min_{x>0}f(x)>0$.

Differentiate and set to $0$: $$f'(x)=(r+1)x^r-\sum_{i=0}^{r-1}{x^i}=0$$ Clearly $f'(t)>0$ for $t\geq 1$, so solution(s) for $f'(x)=0$ lies in $(0,1)$ (at least one exists since $f'(0)<0$ and $f'$is continuous). Use $x^r=\sum_{i=0}^{r-1}{x^i}/(r+1)$, $x<1$ and telescoping product for $\log$:

$$f(x)=x^{r+1}+\log(r+1)-\sum_{i=1}^r{x^i/i}=x\cdot(\sum_{i=0}^{r-1}{x^i}/(r+1))+\log(r+1)-\sum_{i=1}^r{x^i/i}=\sum_{i=1}^{r}x^i(1/(r+1)-1/i)+\log(\prod_{i=1}^r\frac{1+i}{i})>\sum_{i=1}^{r}(1/(r+1)-1/i)+\sum_{i=1}^r\log(1+1/i)=\frac{r}{r+1}-\sum_{i=1}^r\big(1/i-\log(1+1/i)\big)>\frac{r}{r+1}-\gamma>0$$ for $r\geq 2$ (where $\gamma\approx0.577$ is Euler–Mascheroni constant). Separately check the case $r=1$ to get $f(x)\geq \log(2)-1/4>0$.

• How did you get $x^r=\sum_{i=0}^{r-1}{x^i}/(r+1)$ ? Sep 28, 2015 at 14:59
• Rearranging $f'(x)=0$
– A.S.
Sep 28, 2015 at 15:04
• By noting that $x<1$ and $1/(1+r)-1/i<0$.
– A.S.
Sep 28, 2015 at 15:16
• Thanks for your reply -- it's a very nice answer! Sep 28, 2015 at 17:17