Inequality satisfied by real sequence

Question

Question I can't figure out, not for a class just mad I can't get it!

Given:

$$ I_n := \frac{1}{e}\int_0^1{x^ne^xdx} $$

Prove: $$ \frac{1}{n+3} < I_{n+1} < I_n < \frac{1}{n+1} $$

Thought it would be an easy proof but struggling. Thanks!

Answer 1

We have that $I_n = \frac{1}{e} \int \limits_{0}^{1} x^n e^x dx = \int \limits_{0}^{1} x^n e^{x-1} dx$,

Also $x+\frac{(x-1)^2}{3} \leq e^{x-1} \leq x+\frac{(x-1)^2}{2}$ for all $0\leq x \leq 1$.

So $I_n \leq \int \limits_{0}^{1} x^n(x+\frac{(x-1)^2}{2})dx$

Evaluating the integration we get that $I_n < \frac{n+2}{n^2+4 n+3} < \frac{1}{n+1}$ for all $n> -1$.

And $\int \limits_{0}^{1} x^{n+1} (x+\frac{(x-1)^2}{3}) dx \leq I_{n+1}$

Evaluating the integration we get that $\frac{1}{n+3} \leq \frac{1}{3} \left(\frac{1}{n+3}+\frac{1}{n+4}+\frac{1}{n+2}\right) \leq I_{n+1}$ for all $n>-2$.

Done.

Answer 2

From integration by parts we find a recursion that may be rewritten as $I_n = \frac{1-I_{n+1}}{n+1}$. Applying this formula twice:

\begin{align}I_{n+1} &= \frac{1 - I_{n+2}}{n+2} = \frac{1}{n+2} - \frac{1}{n+2} \frac{1 - I_{n+3}}{n+3}\\ &= \frac{1}{n+2}-\frac{1}{(n+2)(n+3)} + \underbrace{\frac{I_{n+3}}{n+3}}_{\geq\, 0}\\ &\geq \frac{1}{n+2}-\frac{1}{(n+2)(n+3)}\\ &= \frac{1}{n+3} \end{align} where the fact $I_{n+3} \geq 0$ follows because $x^{n+3}e^x \geq 0$ for $0\leq x\leq 1$.

Given that $0\leq x\leq 1$, $x^{n+1} = x^n \cdot x \leq x^n$, so $$I_{n+1} = \frac1e \int_0^1 x^{n+1} e^x\;dx \leq \frac1e \int_0^1 x^n e^x = I_n$$

Finally, $e^x \leq e$ for $0\leq x\leq 1$, so $$ I_n = \frac1e \int_0^1 x^n e^x \leq \frac1e \int_0^1 x^n (e) \;dx = \left. \frac{x^{n+1}}{n+1}\right|_0^1 = \frac{1}{n+1}$$