# Thread: Stuff with limits and integrals.. the usual

1. ## Stuff with limits and integrals.. the usual

First prove the inequality
$e^{-t}\Big(1-\frac{e}{2n}t^2\Big)\leq\Big(1-\frac{t}{n}\Big)^n\leq e^{-t}$

for some reason this has been giving me problems, any idea would be appreciated

Then they ask the following but I am not sure how to justify what I am doing to show it:

Show
$\lim_{n\rightarrow\infty}\int_0^{n}\Big(1-\frac{t}{n}\Big)^{n} t^{x-1}dt= \int_0^\infty e^{-t}t^{x-1}dt$
If $0 you must consider a special case neat t=0

Show:
$\int_0^\infty e^{-t}t^{x-1}dt = \lim_{n\rightarrow\infty} \frac{n^xn!}{x(x+1)(x+2)\cdot\cdot\cdot(x+n)}$ for x>0

Any ideas are appreciated, thanks

2. Let's start fron the Newton expansion...

$(1-\frac{t}{n})^{n}= \sum_{k=0}^{n} (-1)^{k}\cdot \binom{n}{k}\cdot (\frac{t}{n})^{k}$ (1)

... that can be rearranged as...

$(1-\frac{t}{n})^{n}= 1-t+\sum_{k=2}^{n}(-1)^{k}\cdot a_{n,k}\cdot \frac{t^{k}}{k!}$ (2)

... where...

$a_{n,k}= \prod_{i=1}^{k-1}(1-\frac{i}{n})$ (3)

It is easy to verify from (3) that is...

$a_{n,k}> 0$ $\forall n>k$

$a_{n+1,k}>a_{n,k}$ $\forall n>k$

$\lim_{n\rightarrow \infty} a_{n,k}=1$ $\forall k$ (4)

From the last of (4) and (2) we derive the well known result...

$\lim_{n\rightarrow \infty} (1-\frac{t}{n})^{n}= \lim_{n\rightarrow \infty} 1-t+\sum_{k=2}^{n}(-1)^{k}\cdot a_{n,k}\cdot \frac{t^{k}}{k!} = \sum_{k=0}^{\infty}(-1)^{k}\frac{t^{k}}{k!}= e^{-t}$ (5)

Now we will use (2) for computing the difference between two consecutive terms of the sequence $(1-\frac{t}{n})^{n}$ ...

$(1-\frac{t}{n+1})^{n+1} - (1-\frac{t}{n})^{n} = \sum_{k=2}^{n} (-1)^{k} \cdot (a_{n+1,k}-a_{n,k}) \cdot \frac{t^{k}}{k!} + (-1)^{n+1}\cdot a_{n+1,n+1}\cdot \frac{t^{n+1}}{(n+1)!}$ (6)

Since the (6) is a series with alternate decreasing terms and the first is positive, the sequence $(1-\frac{t}{n})^{n}$ is increasing and, taking into account (5), we conclude that...

$(1-\frac{t}{n})^{n} $\forall n$ (7)

Kind regards

$\chi$ $\sigma$

3. Post with errors!... please delete!... sorry!...

4. Now we proceed and the first step is computing the inverse of $(1-\frac{t}{n})^{n}$...

$(1-\frac{t}{n})^{-n}= 1 + t + (1+\frac{1}{n})\cdot \frac{t^{2}}{2!} + (1+\frac{1}{n})\cdot(1+\frac{2}{n})\cdot \frac {t^{3}}{3!} +\dots$

$\dots + (1+\frac{1}{n})\cdot(1+\frac{2}{n})\dots (1+\frac{k-1}{n})\cdot \frac {t^{k}}{k!} +\dots$ (1)

... and after the inverse of $e^{-t}\cdot (1-\frac{e}{2n}\cdot t^{2})$...

$\frac{e^{t}}{(1-\frac{e}{2n}\cdot t^{2})} = e^{t}\cdot \{1 + (\frac{e}{2n})\cdot t^{2} + (\frac{e}{2n})^{2}\cdot t^{4} + \dots (\frac{e}{2n})^{k}\cdot t^{2k} + \dots\} =$

$= 1 + t + \{(\frac{e}{2n}) + \frac{1}{2!}\}\cdot t^{2} + \{(\frac{e}{2n}) + \frac{1}{3!}\}\cdot t^{3} + \{(\frac{e}{2n})\cdot \frac{1}{2!} + (\frac{e}{2n})^{2} + \frac{1}{4!}\}\cdot t^{4} + \dots$ (2)

Now we observe in (1) and (2) the coefficients of the term $t^{k}$. For k=0 and k=1 they are equal, for k=2 and $\forall n>0$ is...

$\frac{e}{2n} + \frac{1}{2} > \frac{1}{2n} + \frac{1}{2}$ (3)

... so that for $-t_{0} and $\forall n>0$ is...

$\frac{e^{t}}{(1-\frac{e}{2n}\cdot t^{2})} > (1-\frac{t}{n})^{-n} \rightarrow e^{-t}\cdot (1-\frac{e}{2n}\cdot t^{2}) < (1-\frac{t}{n})^{n}$ (4)

Kind regards

$\chi$ $\sigma$

5. Originally Posted by KZA459
First prove the inequality
$e^{-t}\Big(1-\frac{e}{2n}t^2\Big)\leq\Big(1-\frac{t}{n}\Big)^n\leq e^{-t}$
I think that for both inequalities you will need to assume that t<n. Otherwise, they may well fail. (If t = 2n then the right-hand inequality will fail for n even, and the left-hand one will fail for n odd.)

For the right-hand inequality, take logs. You then need to show that $\ln\bigl(1-\tfrac tn\bigr)\leqslant-\tfrac tn$. But the inequality $\ln(1+x)\leqslant x$ holds for all x>–1 (easy proof by calculus).

The left-hand inequality looks more delicate. But the logarithmic approach might still work there.

6. thanks

7. The double inequality...

$e^{-t}\cdot(1-\frac{e}{2n}\cdot t^{2}) < (1-\frac{t}{n})^{n} < e^{-t}$ (1)

... has been proved [even if with some difficulties ...] in posts #2 and #4. What does matter however is that (1) holds only for $0 where $t_{0}$ must be found and the reason is evident if you consider that...

$\lim_{t \rightarrow +\infty} e^{-t}\cdot(1-\frac{e}{2n}\cdot t^{2}) = \lim_{t \rightarrow +\infty} e^{-t}=0$, $\forall n>0$

... and...

$\lim_{t \rightarrow +\infty} (1-\frac{t}{n})^{n} = + \infty$, n even, $= -\infty$, n odd

In post #2 has been demonstrated that...

$\lim_{n \rightarrow \infty} (1-\frac{t}{n})^{n}= e^{-t}$, $\forall t$ (2)

... and that validates the expression...

$\lim_{n \rightarrow \infty} \int_{0}^{n} (1-\frac{t}{n})^{n} \cdot t^{x-1}\cdot dt= \int_{0}^{\infty} t^{x-1}\cdot e^{-t}\cdot dt$ (3)

The last question will take to me a little more time ...

Kind regards

$\chi$ $\sigma$

8. In post #2 it has been found the expansion...

$(1-\frac{t}{n})^{n} = 1 - t +\sum_{k=2}^{n} (-1)^{k}\cdot a_{n,k}\cdot \frac{t^{k}}{k!}$ (1)

... where...

$a_{n,k}= \prod_{i=1}^{k-1}(1-\frac{i}{n})$ (2)

... and also that is...

$\lim_{n\rightarrow \infty} a_{n,k}= 1$ , $\forall k$ (3)

Using (1) we obtain...

$(1-\frac{t}{n})^{n}\cdot t^{x-1} = t^{x-1} - t^{x} + \sum_{k=2}^{n} (-1)^{k}\cdot a_{n,k}\cdot \frac{t^{x+k-1}}{k!}$ (4)

... so that...

$\int_{0}^{n}(1-\frac{t}{n})^{n}\cdot t^{x-1}\cdot dx = \frac{n^{x}}{x} - \frac{n^{x+1}}{x+1} + \sum_{k=2}^{n}(-1)^{k}\cdot a_{n,k}\cdot \frac{n^{x+k}}{(x+k)\cdot k!}=$

$= n^{x}\cdot \{\frac {1}{x} - \frac{n}{x+1} + \sum_{k=2}^{n}(-1)^{k}\cdot a_{n,k}\cdot \frac{n^{k}}{(x+k)\cdot k!}\}$ (5)

In post #4 it has been demonstrated that...

$\int_{0}^{\infty}t^{x-1}\cdot e^{-t}\cdot dt = \lim_{n\rightarrow \infty} \int_{0}^{n} (1-\frac{t}{n})^{n}\cdot t^{x-1}\cdot dt$ (6)

... so that, taking into account (3) and (5), is...

$\int_{0}^{\infty}t^{x-1}\cdot e^{-t}\cdot dt = \lim_{n\rightarrow \infty} n^{x}\cdot \sum_{k=0}^{n}(-1)^{k}\cdot \frac{n^{k}}{(x+k)\cdot k!}$ (7)

Now we have to search if (7) can be written in different way...

Kind regards

$\chi$ $\sigma$