2
$\begingroup$

If $\displaystyle u_{n}=\int_{0}^{1} t^n \sqrt{t+1} dt$, where $n\geq 1$, prove that $$\frac{\sqrt{2}}{n+1} - \frac{1}{2n^2} \leq u_n \leq \frac{\sqrt{2}}{n+1}.$$

I have already proven the second inequality, but couldn't the first one. I managed to show that $\dfrac{1}{n+1} \leq u_n$, and I wrote a recursive formula by using the $\displaystyle \int_{0}^{1} \dfrac{t^n}{\sqrt{t+1}} dt$ integral, it got worse. Seems to be some easier way.

$\endgroup$
2
$\begingroup$

Since $t\to\sqrt{t+1}$ is concave, it follows that, for $t\in [0,1]$, $$1+(\sqrt{2}-1)t\leq \sqrt{t+1}\leq \sqrt{2}.$$ Hence $$\frac{1}{n+1}+\frac{\sqrt{2}-1}{n+2}\leq \int_{0}^{1} t^n \sqrt{t+1} dt\leq \frac{\sqrt{2}}{n+1}.$$ It remains to show that $$ \frac{\sqrt{2}}{n+1} - \frac{1}{2n^2}\leq \frac{1}{n+1}+\frac{\sqrt{2}-1}{n+2}$$ that is $$\frac{\sqrt{2}-1}{(n+1)(n+2)}\leq \frac{1/2}{n^2}$$ which holds for all $n\geq 1$.

$\endgroup$
0
0
$\begingroup$

$$\int_{0}^{1} t^n \sqrt{t+1} dt\leq\int_{0}^{1} t^n \sqrt{1+1} dt=\frac{\sqrt{2}}{n+1}$$ For the left hand side, with Beta function \begin{align} \int_{0}^{1} t^n (\sqrt{2}-\sqrt{t+1}) dt &= \int_{0}^{1} t^n \dfrac{1-t}{\sqrt{2}+\sqrt{t+1}}\ dt \\ &\leq \dfrac12\int_{0}^{1} t^n(1-t)\ dt \\ &= \frac12\beta(n+1,2) \\ &= \dfrac{1}{2(n+1)(n+2)}\leq\dfrac{1}{2n^2} \end{align}

$\endgroup$
0
$\begingroup$

$\sqrt{t+1}$ is approximately constant on $(0,1)$, hence the following application of Cauchy-Schwarz

$$ I_n=\int_{0}^{1}t^n\sqrt{1+t}\,dt\leq \sqrt{\int_{0}^{1}t^{n}\,dt\int_{0}^{1}t^n(1+t)\,dt}=\color{red}{\frac{1}{n+1}\sqrt{\frac{2n+3}{n+2}}}<\frac{\sqrt{2}}{n+1}$$ is expected to provide a tight upper bound. Indeed, $$ \frac{\sqrt{2}}{n+1}-I_n = \int_{0}^{1}t^n\left(\sqrt{2}-\sqrt{1+t}\right)\,dt = \int_{0}^{1}(1-t)^n\left(\sqrt{2}-\sqrt{2-t}\right)\,dt $$ leads to $$\begin{eqnarray*}\frac{\sqrt{2}}{n+1}-I_n = \int_{0}^{1}\frac{t(1-t)^n}{\sqrt{2}+\sqrt{2-t}}\,dt&\leq&\frac{1}{\sqrt{2}+1}\int_{0}^{1}t(1-t)^n\,dt\\&=&\frac{1}{\sqrt{2}+1}\cdot\frac{1}{(n+1)(n+2)}\\&<&\frac{1}{2(n+1)^2}.\end{eqnarray*}$$ Actually we have $$ I_n = \frac{\sqrt{2}}{n+1}+\frac{C_n}{(n+1)^2} $$ where $C_n\in\left[\frac{1}{2\sqrt{2}},\frac{1}{\sqrt{2}+1}\right]$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.