1. mathematical induction problem

Hi

I am trying to prove the following by mathematical induction.

$\displaystyle 2\sum_{k=1}^n \;\; \frac{1}{k^3}< \;\; \left( 3-\frac{1}{n^2} \right)$

$\displaystyle \forall \; n \;\;\geqslant 2$

first step is to test this for n=2, which I did. If we call above statement P(n) , then
P(2) is true. So we assume that let the statement be true for some
$\displaystyle m\;\in N$ with $\displaystyle m>2$ and the statement is also true for
$\displaystyle i=2,3, \cdots ,m$ . So we have

$\displaystyle 2\sum_{k=1}^m \;\; \frac{1}{k^3}< \;\; \left( 3-\frac{1}{m^2} \right)\cdots (1)$

and such statements from i=2 to i=m. We have to prove that the statement is true for P(m+1). i.e.

$\displaystyle 2\sum_{k=1}^{m+1} \;\; \frac{1}{k^3}< \;\; \left( 3-\frac{1}{(m+1)^2} \right)$

So consider

$\displaystyle 2\sum_{k=1}^{m+1} \;\frac{1}{k^3}$

$\displaystyle =2\sum_{k=1}^m \;\frac{1}{k^3}+\frac{2}{(m+1)^3}$

Now we can use equation 1 here

$\displaystyle < \;\left( 3-\frac{1}{m^2}\right) +\frac{2}{(m+1)^3}$

I tried to manipulate the second term here. We see that for m > 2 , we have

$\displaystyle (m+1)^3 > 2(m+1)^2}$

$\displaystyle \Rightarrow \; \frac{1}{(m+1)^3} < \frac{1}{2(m+1)^2}$

$\displaystyle \Rightarrow \; \frac{2}{(m+1)^3} < \frac{1}{(m+1)^2}$

so using this above, we have

$\displaystyle 2\sum_{k=1}^{m+1} \;\frac{1}{k^3} < \left( 3-\frac{1}{m^2} \right)+\frac{1}{(m+1)^2}$

I don't know where to go from here....... can anybody suggest something ?

thanks

2. The last equation should be

$\displaystyle 2\sum_{k=1}^{m+1} \;\frac{1}{k^3} < \left( 3-\frac{1}{m^2} \right)+\frac{1}{(m+1)^2}$

for some reasons the Edit button didn't work properly. It couldn't save the changed matter. So I am replying to my own post.

3. I would try

P(m)

$\displaystyle 2\sum_{k=1}^m\frac{1}{k^3}<3-\frac{1}{m^2}$

P(m+1)

$\displaystyle 2\sum_{k=1}^{m+1}\frac{1}{k^3}<3-\frac{1}{(m+1)^2}$

Then if P(m) is true

$\displaystyle 2\sum_{k=1}^{m+1}\frac{1}{k^3}<3-\frac{1}{m^2}+\frac{2}{(m+1)^3}$

Then P(m+1) will DEFINATELY be true if

$\displaystyle 3-\frac{1}{(m+1)^2}>3-\frac{1}{m^2}+\frac{2}{(m+1)^3}$

Then to prove...

$\displaystyle \frac{2}{(m+1)^3}+\frac{1}{(m+1)^2}-\frac{1}{m^2}<0\;\;\;?$

$\displaystyle \frac{2+m+1}{(m+1)^3}<\frac{1}{m^2}\;\;\;?$

$\displaystyle \frac{3m^2+m^3}{(m+1)^3}<1\;\;\;?$

which is true if you expand the denominator.

4. Originally Posted by issacnewton
The last equation should be

$\displaystyle 2\sum_{k=1}^{m+1} \;\frac{1}{k^3} < \left( 3-\frac{1}{m^2} \right)+\frac{1}{(m+1)^2}$

for some reasons the Edit button didn't work properly. It couldn't save the changed matter. So I am replying to my own post.
For some reason editing only works if you "Go Advanced".

5. Archie, thanks .... but this is not the usual way its done.. I spent lot of time thinking about this....

6. Originally Posted by issacnewton
Hi

I am trying to prove the following by mathematical induction.

$\displaystyle 2\sum_{k=1}^n \;\; \frac{1}{k^3}< \;\; \left( 3-\frac{1}{n^2} \right)$

$\displaystyle \forall \; n \;\;\geqslant 2$

first step is to test this for n=2, which I did. If we call above statement P(n) , then
P(2) is true. So we assume that let the statement be true for some
$\displaystyle m\;\in N$ with $\displaystyle m>2$ and the statement is also true for
$\displaystyle i=2,3, \cdots ,m$ . So we have

$\displaystyle 2\sum_{k=1}^m \;\; \frac{1}{k^3}< \;\; \left( 3-\frac{1}{m^2} \right)\cdots (1)$

and such statements from i=2 to i=m. We have to prove that the statement is true for P(m+1). i.e.

$\displaystyle 2\sum_{k=1}^{m+1} \;\; \frac{1}{k^3}< \;\; \left( 3-\frac{1}{(m+1)^2} \right)$

So consider

$\displaystyle 2\sum_{k=1}^{m+1} \;\frac{1}{k^3}$

$\displaystyle =2\sum_{k=1}^m \;\frac{1}{k^3}+\frac{2}{(m+1)^3}$

Now we can use equation 1 here

$\displaystyle < \;\left( 3-\frac{1}{m^2}\right) +\frac{2}{(m+1)^3}$

I tried to manipulate the second term here. We see that for m > 2 , we have

$\displaystyle (m+1)^3 > 2(m+1)^2}$

$\displaystyle \Rightarrow \; \frac{1}{(m+1)^3} < \frac{1}{2(m+1)^2}$

$\displaystyle \Rightarrow \; \frac{2}{(m+1)^3} < \frac{1}{(m+1)^2}$

so using this above, we have

$\displaystyle 2\sum_{k=1}^{m+1} \;\frac{1}{k^3} < \left( 3-\frac{1}{m^2} \right)+\frac{1}{(m+1)^2}$

I don't know where to go from here....... can anybody suggest something ?

thanks
From here, can you see you want to show that P(m+1) is absolutely true if

$\displaystyle 3-\frac{1}{m^2}+\frac{1}{(m+1)^2}>3-\frac{1}{(m+1)^2}\;\;\;?$

Then

$\displaystyle \frac{2}{(m+1)^2}>\frac{1}{m^2}$

which can be written as

$\displaystyle m^2>2m+1\;\;?$

which is true for m=3 and above.

7. I did understand your original reasoning too. But this one is nicer too. Many problems in analysis basically boil down to playing with inequalities I guess