How to find a meaningful bound on a sequence that is known to go to $0$ 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.
 A: 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 $\sum_k I_k$.)
A: (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, $1 - kI_{k-1}=I_k \lt \cfrac{1}{k+1} \implies k I_{k-1} \gt 1  - \cfrac{1}{k+1}=\cfrac{k}{k+1}\implies I_{k-1} \gt \cfrac{1}{k+1}$
A: $$I_k = \int_0^1 x^ke^{x-1} \, dx$$
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.
