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