How to find a meaningful bound on a sequence that is known to go to $0$

Question

I am doing a programming exercise (quite an interesting one, actually) on the sequence $\{I_k\}_{k\in\Bbb{N}}$

$$I_k = \int_0^1 x^ke^{x-1} dx$$

and I am also given a - already proven by me - recurrence formula,

$$I_k = 1 - kI_{k-1}$$

and at some point I must show that the $I_k$ go to $0$. I was trying to do it just with the recurrence formula and the fact that $I_1 = \frac1e$, with induction. I tried proving $I_k < \frac{1}{k+1}$ but in proving it I ended up needing to assume that $\frac{1}{k+1} < I_{k-1}$. So basically I wanted to prove

$$\frac{1}{k+2} < I_k < \frac{1}{k+1} \tag{1}$$

I verified it for $k$ up to $10$ so it does feel like it should be true. With induction, proving $I_k < \frac{1}{k+1}$ assuming (1), is trivial. However, when I try to prove $\frac{1}{k+2} < I_k$ I end up with

$$I_{k-1} < \frac{1}{k}\frac{k+1}{k+2}$$

which made me write

$$I_{k-1} < \frac{1}{k}\frac{k+1}{k+2} < \frac{1}{k}, \text{true, by the induction hypothesis}$$

However, I can't do this. I know $I_{k-1}< \frac{1}{k}$ but that does not let me write what I wrote nor prove what I want to prove. Is there anyone out there able to lend me a hand?

The ideal would be to use induction to prove (1).

If I fail to do that, the second ideal thing would be to use induction to prove that the $I_k$, starting at some $k_0$, are bounded by something that does not increase. It can even be a constant. For example, showing $I_k < 1\ \forall k$ would be good enough for my purposes.

Also, if anyone knows if this sequence (or these integrals) has any name, I would be able to better search for things that could help.

Answer 1

(Expanding on my previously posted comments.)

I am also given a - already proven by me - recurrence formula,

$$I_k = 1 - kI_{k-1}$$

Once the recurrence relation is determined, the bounds follow directly from just the positivity of the integrand $\,x^k e^{x-1} \gt 0\,$ for $\,x \gt 0\,$, which immediately implies $\,I_k \gt 0\,$ for all $\forall k \in \mathbb{N}\,$. Therefore:

• $1 - kI_{k-1}=I_k \gt 0 \implies k I_{k-1} \lt 1 \implies I_{k-1} \lt \cfrac{1}{k} \quad\quad$

• then, $1 - kI_{k-1}=I_k \lt \cfrac{1}{k+1} \implies k I_{k-1} \gt 1 - \cfrac{1}{k+1}=\cfrac{k}{k+1}\implies I_{k-1} \gt \cfrac{1}{k+1}$

Answer 2

It is not clear to me whether the question suggests induction or requires it. If induction is merely suggested, then here is an answer which declines the suggestion!

Majorize via $$0 \le x^ke^{x-1} \le x^k\:\ \mbox{for all}\ x\in[0,1].$$ By this, we have that $$ 0 \le I_k = \int_0^1 x^k e^{x-1}\,dx \le \int_0^1 x^k \, dx = \frac 1 {k+1} \to 0 \text{ as } k\to\infty. $$ which I believe does what you want.

(This would fail, of course, if you needed $\sum_k I_k$ to be bounded—but that isn't a fault of the proof but rather of the series. In any case, you have not asked for boundedness in $\sum_k I_k$.)

Answer 3

$$I_k = \int_0^1 x^ke^{x-1} \, dx$$

Directly from the integral:

$$\frac1{e} \lt e^{x-1} \lt 1 $$ so $$\frac1{e(k+1)} \lt I_k \lt \frac1{k+1}. $$

Since $e^{x-1} > x$ (because $e^z > 1+z$), $I_k > \int_0^1 x^kx \,dx =\frac1{k+2} $. This is what you wanted.

Trying to be more precise:

Interpolating at $0$ and $1$, since $(e^{x-1})'' > 0$, $e^{x-1} \lt (1-\frac1{e})x+\frac1{e} $, so

\begin{align} I_k &\lt \int_0^1 x^k \left(\left(1-\frac1 e\right)x+\frac 1 e\right) \, dx\\ &=\left(1-\frac 1 e\right)\frac1{k+2}+\frac1{e(k+1)}\\ &=\frac1{k+2}+\frac1{e(k+1)(k+2)} \end{align}

so that $0 \lt I_k-\frac1{k+2} \lt \frac1{e(k+1)(k+2)} $.

Let $d(x) =(1-\frac1{e})x+\frac1{e}-e^{x-1} =(1-\frac1{e})x+\frac1{e}(1-e^{x}) $. $d(0) = d(1) = 0$ and $d'(x) =(1-\frac1{e})-e^{x-1} =0 $ when

\begin{align} x &=x_0\\ &=1+\ln(1-\frac1{e})\\ &=1+\ln(e-1)-1\\ &=\ln(e-1)\\ &\approx 0.54132\\ \end{align}

where $d(x_0) = \frac1{e}(2 - e + (e - 1) \ln(e - 1)) \approx 0.077941 $.

Therefore $e^{x-1} \ge d(x)-d(x_0) $ so $I_k \ge \frac1{k+1}-(1-\frac1{e})\frac1{(k+1)(k+2)}-d(x_0) $.

That's enough for now.