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Prove that $$\int\limits_0^1\frac{x^{n+1}}{x+1}dx<\frac{1}{2(n+1)}$$

On simplifying by parts we get:

$$\int\limits_0^1\frac{x^{n+1}}{x+1}dx=\frac{1}{2(n+2)}+\int\limits_0^1\frac{x^{n+2}}{(x+1)^2(n+2)}dx$$

Thus if we prove that$$\max\left(\displaystyle\int\limits_0^1\frac{x^{n+2}}{(x+1)^2(n+2)}dx\right)<\frac{1}{2(n+1)(n+2)}$$

We will be able to prove the above inequality

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  • $\begingroup$ Just to clarify, are you looking to prove the bound in your post or the bound in your title? $\endgroup$ Commented Sep 9, 2021 at 6:47
  • $\begingroup$ @AlannRosas I want to prove the bound in my title $\endgroup$
    – Tatai
    Commented Sep 9, 2021 at 6:49

3 Answers 3

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Simply observe that for $x \in [0,1]$ and $n \in \mathbb Z^+$, $$0 \le (1-x)x^n = (x+1)x^n - 2x^{n+1},$$ the first inequality arising from the fact that $x \ge 0$ and $1-x \ge 0$. Consequently, $$0 \le \frac{x^{n+1}}{x+1} \le \frac{x^n}{2},$$ and integrating gives the desired inequality.

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    $\begingroup$ That's simpler. – One could also directly observe that $\frac{x}{x+1} \le \frac 12$ because the expression is increasing on $[0, 1]$. $\endgroup$
    – Martin R
    Commented Sep 9, 2021 at 7:00
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$$\int_0^1 \frac{x^{n+1}}{x+1}\le \frac{1}{2}\int_0^1\frac{x^{n+1}+x^{n}}{x+1}=\frac{1}{2(n+1)}$$ here we used when $x\in [0,1] $ $x^{n+1}\le x^n$

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    $\begingroup$ Nice one. You're a wizard, Harry! $\endgroup$ Commented Sep 9, 2021 at 7:01
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Inspired by the other answers: $f(x) = \frac{x}{1+x}$ is concave on $[0, \infty)$, and therefore $$ f(x) \le f(1) + (x - 1) f'(1) = \frac 1 4(1+x) $$ for $x \ge 0$. It follows that $$ \int_0^1 \frac{x^n}{1+x}\, dx \le \frac 1 4 \int_0^1 \left(x^{n+1} + x^{n+2} \right) \, dx \\ = \frac 1 4 \left( \frac 1{n+2} + \frac 1{n+3}\right) < \frac{1}{2(n+2)} \, , $$ which is slightly sharper than the desired estimate.

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