Proving an equality using induction

[jetpac]

Proving an inequality using induction

Homework Statement

Hey, sorry for any bad formatting, I did my best..

Prove for all positive integers n:

<br /> \sum_{i = 1}^n \frac{2i^2 - 1}{i^4} \leq 4 - ( \frac{2n+1}{n^2} )<br />

The Attempt at a Solution

First prove for n=1:

<br /> \frac{2(1)-1}{1} = 1<br />
<br /> 4 - \frac{2+1}{1} = 1 <br />

therefore it holds for n = 1

Then assuming the equation holds for any positive integer m, does it hold for n = m+1:

n = m:
<br /> \sum_{i = 1}^m \frac{2i^2 - 1}{i^4} \leq 4 - ( \frac{2m+1}{m^2} )<br />

n = m+1:

Subbing m+1 into the RHS gives:

<br /> 4 - \frac{2m + 3}{(m+1)^2}<br />


However subbing (i=m+1) into the LHS and adding it to the RHS with (n=m) gives:

<br /> (4 - \frac{2m+1}{m^2} ) + \frac{2(m+1)^2 - 1}{(m+1)^4}<br /> <br /> = 4 - (\frac{2m+1}{m^2} - \frac{2(m+1)^2 - 1}{(m+1)^4})<br /> <br /> = 4 - (\frac{(2m + 1)(m + 1)^4-m^2(2m^2 + 4m + 1)}{m^2(m+1)^4})<br />


Next multiply the first equation top and bottom by <br /> m^2(m+1)^2 gives:

<br /> 4 - \frac{2m + 3}{(m+1)^2} * \frac{m^2(m+1)^2}{m^2(m+1)^2}<br /> = 4 - (\frac{m^2(2m+3)(m+1)^2}{m^2(m+1)^4})<br />


So essentially I sub (i = m+1) into the LHS and (n=m) into the RHS and add them (giving the next value in the sequence). Then I compare this to what happens when I just sub (n = m+1) directly into the RHS and isolate exactly what differs between the two methods. The result is that you get 4 - \frac{some number}{m^2(m+1)^4} in both cases. The minimum value of {some number} can be found by subbing m=1 into it (since m is positive). So then if {some number} on the LHS is greater than {some number} on the RHS then the original statement also holds for n = m + 1 and therefore all positive integers m?

 

[Mark44]

Homework Statement

Hey, sorry for any bad formatting, I did my best..

Prove for all positive integers n:

\sum (2i^2 - 1)/i^4 \leq 4 - ((2n+1) / n^2 )

The Attempt at a Solution

First prove for n=1:

2(1)-1/1 = 1 (LHS)
4 - (2+1/1) = 1 (RHS)

therefore it holds for n = 1

Then assuming it holds for some pos int m, does it hold for n = m+1:

Subbing m+1 into the RHS gives:

4 - (2m + 3)/(m+1)^2

However adding (n=m+1) on the LHS to the RHS (n=m):

(4 - (2m+1)/(m^2) ) + (2(m+1)^2 - 1)/(m+1)^4

= 4 - ( 2m+1/m^2 - (2(m+1)^2 - 1)/(m+1)^4 )

So if ( 2m+1/m^2 - (2(m+1)^2 - 1)/(m+1)^4 ) \geq (2m + 3)/(m+1)^2

Then it will hold so put manipulate both sides to have the same denominator:

LHS = ( m^2(2m + 3)(m + 1)^2 ) / (m^2(m + 1)^4)

RHS = ( (2m+1)(m+1)^4 - m^2(2m^2 + 4m + 1) ) / (m^2(m + 1)^4)


Since m >= 1 then subbing m=1 into the following equations:

( (2m+1)(m+1)^4 - m^2(2m^2 + 4m + 1) ) \geq ( m^2(2m + 3)(m + 1)^2 )

gives LHS >= 41, RHS >= 20

And 4 - 41/((m^2(m + 1)^4)) < 4 - 21/((m^2(m + 1)^4))

And so it holds for n = m+1 and by the process of induction all positive integer values of m.

It's difficult for me to follow what you have done, especially since you left off a very important step.

Then assuming it holds for some pos int m, does it hold for n = m+1

What is the statement that you are assuming to be true when n = m?

Also, here's some help with the LaTeX. Click on this expression to see what I did.
\sum_{i = 1}^n (2i^2 - 1)/i^4 \leq 4 - ((2n+1) / n^2 )

 

[jetpac]

You're right it was pretty messy. I've edited the original post - let me know if that's any easier to understand! Thanks for your help!

 

[Mark44]

That's much easier to follow. I don't have time for a close look now, but will take a look in about an hour, unless someone else jumps in.

 

[jetpac]

Great thanks, if you get a chance at all today or tomorrow I'd appreciate it a lot :)

 

[jetpac]

Anyone able to help me out on this one?

 

[Mark44]

Your induction hypothesis is
\sum_{i = 1}^m (2i^2 - 1)/i^4 \leq 4 - ((2m+1) / m^2 )

You need to use this assumption to show that
\sum_{i = 1}^{m+1} (2i^2 - 1)/i^4 \leq 4 - ((2(m + 1) +1) / (m + 1)^2 )

Start with this:
\sum_{i = 1}^{m + 1} (2i^2 - 1)/i^4 = \sum_{i = 1}^m (2i^2 - 1)/i^4 + \frac{2(m+1)^2 - 1}{(m + 1)^4}
\leq 4 - \frac{2m + 1}{m^2} + \frac{2(m+1)^2 - 1}{(m + 1)^4}

The first two terms above come from the induction hypothesis. Can you work with this to turn it into the right side of the statement you're trying to prove?

 

[jetpac]

Hey Mark, thanks for taking the time to have a look at it. I think I got as far as that calculation above but it was what to do next that I was sort of having difficulty with.

 

[Mark44]

I've tried a few things without much success. The best advice I can give is to work on the last two terms in the last line of what I have in the last line of my previous post. One of the things I tried that might be fruitful is the idea that
4 - (2m + 1)/m^2 <= 4 - (2m + 1)/(m + 1)^2
The idea here is that we're subtracting a smaller quantity from 4 on the right, which means that the overall expression on the right is larger.