1.
a) Calculate
and 

As we can see the first term cancels out with the fourth, the third with the sixth, and so on and all we are left with is the second and second last terms:


b) Calculate
using the previous result.

Defining
the previous sum can be written as

This last result apparently has a funny story. Mengoli was the first one to calculate
.
At the time this happened people did research in mathematics (I’m using this term rather abusively) in a somewhat different vein. They didn’t rush to print what they found like today.
Many times people held out their results for years while tormenting their rivals about what they found.
This is exactly what Mengoli did. In the times he was around the theory of series wasn’t much developed, thus this result, that we can calculate without being particularly brilliant in Mathematics, was something to take note of.
So, he wrote some letters to people saying that
, but not how he concluded that.
The other mathematicians he sent the result too didn’t know about his methods and all they could do was to add numbers up explicitly and the only thing they could see was that even though they could sum more and more terms the result was always less than
and was got nearer and nearer to
.
Of course this didn’t prove nothing since summing up a billion terms isn’t the same as summing an infinite number of terms and everyone but Mengoli was dumbfounded with that surprising result.
c) Calculate 
In this exercise what we are calculating is the sum of
consecutive odd numbers. This result was already known to the ancient Greeks and the result wasn’t nothing short to astounding to them.
But enough with the talk already:
![{\begin{aligned} \displaystyle \sum_{k=0}^{n-1}(2k+1)&=\sum_{k=0}^{n-1}\left[ (k+1)^2-k^2\right]\\ &= \sum_{k=0}^{n-1}(u_{k+1}-u_k) \end{aligned}} {\begin{aligned} \displaystyle \sum_{k=0}^{n-1}(2k+1)&=\sum_{k=0}^{n-1}\left[ (k+1)^2-k^2\right]\\ &= \sum_{k=0}^{n-1}(u_{k+1}-u_k) \end{aligned}}](http://s0.wp.com/latex.php?latex=%7B%5Cbegin%7Baligned%7D+%5Cdisplaystyle+%5Csum_%7Bk%3D0%7D%5E%7Bn-1%7D%282k%2B1%29%26%3D%5Csum_%7Bk%3D0%7D%5E%7Bn-1%7D%5Cleft%5B+%28k%2B1%29%5E2-k%5E2%5Cright%5D%5C%5C+%26%3D+%5Csum_%7Bk%3D0%7D%5E%7Bn-1%7D%28u_%7Bk%2B1%7D-u_k%29+%5Cend%7Baligned%7D%7D&bg=000000&fg=ffffff&s=0)
With 
Using the now familiar formula

An astounding result indeed!
Just look to
, interpret the result and try not to be as surprised as the ancient Greeks were.
2.
a) Using 1.a) and
calculate 
![{\begin{aligned} \displaystyle \sum_{k=0}^{n-1} a^k &= \displaystyle\sum_{k=0}^{n-1} \left[ a^k\frac{a-1}{a-1}\right]\\ &= \displaystyle \frac{1}{a-1}\sum_{k=0}^{n-1}\left( a^{k+1}-a^k\right)\\ &= \displaystyle\frac{1}{a-1}(a^n-1)\\ &= \displaystyle\frac{a^n-1}{a-1} \end{aligned}} {\begin{aligned} \displaystyle \sum_{k=0}^{n-1} a^k &= \displaystyle\sum_{k=0}^{n-1} \left[ a^k\frac{a-1}{a-1}\right]\\ &= \displaystyle \frac{1}{a-1}\sum_{k=0}^{n-1}\left( a^{k+1}-a^k\right)\\ &= \displaystyle\frac{1}{a-1}(a^n-1)\\ &= \displaystyle\frac{a^n-1}{a-1} \end{aligned}}](http://s0.wp.com/latex.php?latex=%7B%5Cbegin%7Baligned%7D+%5Cdisplaystyle+%5Csum_%7Bk%3D0%7D%5E%7Bn-1%7D+a%5Ek+%26%3D+%5Cdisplaystyle%5Csum_%7Bk%3D0%7D%5E%7Bn-1%7D+%5Cleft%5B+a%5Ek%5Cfrac%7Ba-1%7D%7Ba-1%7D%5Cright%5D%5C%5C+%26%3D+%5Cdisplaystyle+%5Cfrac%7B1%7D%7Ba-1%7D%5Csum_%7Bk%3D0%7D%5E%7Bn-1%7D%5Cleft%28+a%5E%7Bk%2B1%7D-a%5Ek%5Cright%29%5C%5C+%26%3D+%5Cdisplaystyle%5Cfrac%7B1%7D%7Ba-1%7D%28a%5En-1%29%5C%5C+%26%3D+%5Cdisplaystyle%5Cfrac%7Ba%5En-1%7D%7Ba-1%7D+%5Cend%7Baligned%7D%7D&bg=000000&fg=ffffff&s=0)
b) Using a) establish the Bernoulli inequality
if
and 
If
it is
which is trivially true.
If
it is
which is trivially true.
For
and
it is:

Thus

Since 
Finally if
it is

Thus

Since 
c) Use b) to calculate
if
and then conclude that
if
.
By b) it is

Hence 
For the second part of the exercise we will calculate instead
since that we know that 
Let us make a change of variable
. Thus
and

3. Consider the sequences
and 
a) Calculate
and
. Then use Bernoulli’s inequality to show that
is strictly decreasing and that
is strictly increasing.

After having calculated
we can use Bernoulli’s inequality, with
, to conclude that
is strictly decreasing.

Thus
is strictly decreasing.
With a similar technique we can prove that

After that by using Bernoulli’s inequality like in the previous example one can show that
and thus
is strictly increasing.
c) Using a) and b) and
prove the following inequalities
.

We already know that
is decreasing so it is 
On the other hand
is increasing and
so
.
Hence 
d) Use c) to prove that 

And now for the second part of the inequality:

In conclusion it is 
4.
a) Using 3d) show that

From

With a similar reasoning we can also prove that
.
Thus it is 
b) Sum the previous inequalities between
.

Now

And

It also is

And 
Thus it is 
c) Conclude the following inequalities
and establish Stirling's approximation
with 

On the other hand

Thus

And from this follows 
Defining
it is
with 
5.
Show that
and that 
We know that

Thus
and this equivalent to saying that 
Let
. In this case it is
. Since
is a subsequence of
we know that
and so it also is
.
6. Show that
and 
By hypothesis it is
,
with
.
Substituting the second equality in the first we obtain
.
Let
and we write
with
.
Thus 
7. Let
and
. Show that
.
and
with
and
bounded sequences. Now

Let
it is
since
is bounded.
Thus 
8. Using Stirling’s approximation show that 
We know that it is
with
. Thus

Where we used the fact that
is decreasing function.
Thus
is bounded and so
as desired.
Like this:
Like Loading...