Prove Summation Inequality: $\frac{1}{2n-1} > \sum_{k=n}^{2n-2}\frac{1}{k^2}$

In summary, in order to prove the given inequality, it suffices to show that adding -\frac1{n^2} + \frac1{(2n-1)^2} + \frac1{(2n)^2} to the left side of the given summation will result in a value less than or equal to the right side. This can be achieved by using telescoping sums and showing that the resulting inequality holds true for all values of $n\geqslant2$.
  • #1
MarkFL
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Prove the following:

\(\displaystyle \sum_{k=n}^{2n-2}\frac{1}{k^2}<\frac{1}{2n-1}\) where \(\displaystyle 2\le n\)
 
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  • #2
MarkFL said:
Prove the following:

\(\displaystyle \sum_{k=n}^{2n-2}\frac{1}{k^2}<\frac{1}{2n-1}\) where \(\displaystyle 2\le n\)
[sp]\(\displaystyle \sum_{k=n}^{2n-2}\frac{1}{k^2}\) is the lower Riemann sum for the function $1/x^2$ on the interval $[n-1,2n-2]$ using subintervals of length $1$. That is less than the value of the integral of the function over that interval, namely $$\int_{n-1}^{2n-2}\frac1{x^2}\,dx = \left[-\frac1x \right]_{n-1}^{2n-2} = -\frac1{2n-2} + \frac1{n-1} = \frac1{2n-2} < \frac1{2n-1}.$$[/sp]
 
  • #3
Hi Opalg,

At the end of your proof you write $\frac{1}{2n-2} < \frac{1}{2n-1}$, but in fact $\frac{1}{2n-2} > \frac{1}{2n-1}$.
 
  • #4
Euge said:
Hi Opalg,

At the end of your proof you write $\frac{1}{2n-2} < \frac{1}{2n-1}$, but in fact $\frac{1}{2n-2} > \frac{1}{2n-1}$.
Oops, seems like I was having a senior moment there. (Blush)
 
  • #5
MarkFL said:
Prove the following:

\(\displaystyle \sum_{k=n}^{2n-2}\frac{1}{k^2}<\frac{1}{2n-1}\) where \(\displaystyle 2\le n\)
Let's hope this attempt is better than the previous one.
[sp]$k^2>k^2-1 = (k-1)(k+1)$. Therefore \(\displaystyle \frac1{k^2} < \frac1{(k-1)(k+1)} = \frac1{2(k-1)} - \frac1{2(k+1)},\) from which $$\sum_{k=n}^{2n-2}\frac{1}{k^2}< \sum_{k=n}^{2n-2}\Bigl(\frac1{2(k-1)} - \frac1{2(k+1)}\Bigr).$$ The sum on the right telescopes, leaving two orphan terms at the beginning and two at the end, to give $$\sum_{k=n}^{2n-2}\frac{1}{k^2}< \frac1{2(n-1)} + \frac1{2n} - \frac1{2(2n-2)} - \frac1{2(2n-1)} = \frac1{4(n-1)} + \frac1{2n} - \frac1{2(2n-1)}.$$ We want this to be less than (or equal to) \(\displaystyle \frac1{2n-1}\), in other words $$\frac1{4(n-1)} + \frac1{2n} \leqslant \frac3{2(2n-1)}.$$ Multiply through by the denominators to see that this is equivalent to $$n(2n-1) + 2(n-1)(2n-1) \leqslant 6n(n-1).$$ That in turn is equivalent to $$\cancel{6n^2} -7n+2 \leqslant \cancel{6n^2} - 6n,$$ $$-n\leqslant-2.$$ Thus the given inequality holds provided that $n\geqslant2$.[/sp]
 
  • #6
Hi Opalg,

Yes, this proof works! Nice one! (Cool)
 
  • #7
Opalg said:
Let's hope this attempt is better than the previous one.
[sp]$k^2>k^2-1 = (k-1)(k+1)$. Therefore \(\displaystyle \frac1{k^2} < \frac1{(k-1)(k+1)} = \frac1{2(k-1)} - \frac1{2(k+1)},\) from which $$\sum_{k=n}^{2n-2}\frac{1}{k^2}< \sum_{k=n}^{2n-2}\Bigl(\frac1{2(k-1)} - \frac1{2(k+1)}\Bigr).$$ The sum on the right telescopes, leaving two orphan terms at the beginning and two at the end, to give $$\sum_{k=n}^{2n-2}\frac{1}{k^2}< \frac1{2(n-1)} + \frac1{2n} - \frac1{2(2n-2)} - \frac1{2(2n-1)} = \frac1{4(n-1)} + \frac1{2n} - \frac1{2(2n-1)}.$$ We want this to be less than (or equal to) \(\displaystyle \frac1{2n-1}\), in other words $$\frac1{4(n-1)} + \frac1{2n} \leqslant \frac3{2(2n-1)}.$$ Multiply through by the denominators to see that this is equivalent to $$n(2n-1) + 2(n-1)(2n-1) \leqslant 6n(n-1).$$ That in turn is equivalent to $$\cancel{6n^2} -7n+2 \leqslant \cancel{6n^2} - 6n,$$ $$-n\leqslant-2.$$ Thus the given inequality holds provided that $n\geqslant2$.[/sp]

Nicely done, Opalg! (Yes)

Here's my solution:

We are given to prove:

\(\displaystyle \sum_{k=n}^{2n-2}\frac{1}{k^2}<\frac{1}{2n-1}\) where \(\displaystyle 2\le n\)

Using induction, we first check the base case $P_2$:

\(\displaystyle \sum_{k=2}^{2\cdot2-2}\frac{1}{k^2}<\frac{1}{2\cdot2-1}\)

\(\displaystyle \sum_{k=2}^{2}\frac{1}{k^2}<\frac{1}{2\cdot2-1}\)

\(\displaystyle \frac{1}{2^2}<\frac{1}{2^2-1}\)

This is true, since the denominator on the right is smaller and the numerators are equal. Taking what we are given to prove as $P_n$, we see that we want to add to the left:

\(\displaystyle -\frac{1}{n^2}+\left(\frac{1}{2(n+1)-2}\right)^2=\left(\frac{1}{2n}\right)^2-\frac{1}{n^2}=-\frac{3}{4n^2}\)

And we want to add to the right:

\(\displaystyle \frac{1}{2(n+1)-1}-\frac{1}{2n-1}=\frac{1}{2n+1}-\frac{1}{2n-1}=\frac{2n-1-2n-1}{(2n+1)(2n-1)}=-\frac{2}{4n^2-1}\)

So, what we need to show is that:

\(\displaystyle -\frac{3}{4n^2}<-\frac{2}{4n^2-1}\)

If we simultaneously invert both sides and multiply through by -6, the direction of the inequality remains the same, and we have:

\(\displaystyle 8n^2<12n^2-3\)

Subtract through by $8n^2$:

\(\displaystyle 0<4n^2-3\)

This is true for $2\le n$, thus we may add:

\(\displaystyle -\frac{1}{n^2}+\left(\frac{1}{2(n+1)-2}\right)^2<\frac{1}{2(n+1)-1}-\frac{1}{2n-1}\)

to $P_n$:

\(\displaystyle \sum_{k=n}^{2n-2}\frac{1}{k^2}<\frac{1}{2n-1}\)

to get:

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

which then simplifies to:

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

We have derived $P_{n+1}$ from $P_n$ thereby completing the proof by induction.
 
  • #8
[sp]
MarkFL said:
... we want to add to the left:

\(\displaystyle -\frac{1}{n^2}+\left(\frac{1}{2(n+1)-2}\right)^2=\left(\frac{1}{2n}\right)^2-\frac{1}{n^2}=-\frac{3}{4n^2}\)
I think that what you need to add on the left is \(\displaystyle -\frac1{n^2} + \frac1{(2n-1)^2} + \frac1{(2n)^2}\) (Instead of going from $n$ to $2n-2$, the summation will go from $n+1$ to $2n$. So there is one term dropped from the start, but two terms added at the end.) That is why I could not make the proof by induction work.[/sp]
 
  • #9
Opalg said:
[sp]
I think that what you need to add on the left is \(\displaystyle -\frac1{n^2} + \frac1{(2n-1)^2} + \frac1{(2n)^2}\) (Instead of going from $n$ to $2n-2$, the summation will go from $n+1$ to $2n$. So there is one term dropped from the start, but two terms added at the end.) That is why I could not make the proof by induction work.[/sp]

You are absolutely correct...for each increment of $n$ the number of terms in the sum increases by 1. I will try to fix my proof soon. :)
 
  • #10
Because we find:

\(\displaystyle \frac{d}{dx}\left(x^{-2}\right)=-2x^{-3}\)

which is increasing as $x$ increases, we may state:

\(\displaystyle \sum_{k=n}^{2n-2}\frac{1}{k^2}<\int_{n}^{2n-1}\left(x-\frac{1}{2}\right)^{-2}\,dx\)

Hence:

\(\displaystyle \sum_{k=n}^{2n-2}\frac{1}{k^2}<\left.-\frac{1}{x-\dfrac{1}{2}}\right|_{n}^{2n-1}\)

\(\displaystyle \sum_{k=n}^{2n-2}\frac{1}{k^2}<\frac{1}{n-\frac{1}{2}}-\frac{1}{2n-\dfrac{3}{2}}\)

\(\displaystyle \sum_{k=n}^{2n-2}\frac{1}{k^2}<2\left(\frac{1}{2n-1}-\frac{1}{4n-3}\right)\)

\(\displaystyle \sum_{k=n}^{2n-2}\frac{1}{k^2}<2\left(\frac{4n-3-2n+1}{(2n-1)(4n-3)}\right)\)

\(\displaystyle \sum_{k=n}^{2n-2}\frac{1}{k^2}<\frac{4n-4}{4n-3}\cdot\frac{1}{2n-1}<\frac{1}{2n-1}\)
 

FAQ: Prove Summation Inequality: $\frac{1}{2n-1} > \sum_{k=n}^{2n-2}\frac{1}{k^2}$

What is the summation inequality?

The summation inequality is a mathematical expression that compares the sum of a series of terms to the sum of a related series of terms. It is often used to compare the size of a finite sum to an infinite sum.

How do you prove the summation inequality?

To prove the summation inequality, we can use mathematical induction. First, we show that the inequality is true for the base case, n=1. Then, we assume that the inequality holds for a general case, n=k. Finally, we use this assumption to prove that the inequality also holds for the next case, n=k+1. This completes the proof by induction.

What is the significance of the inequality?

The summation inequality is significant because it allows us to compare the size of different series of terms. In this particular case, the inequality helps us to understand the relationship between the sum of the reciprocals of odd numbers and the sum of the reciprocals of squares of numbers.

Can you provide an example of the summation inequality in action?

One example of the summation inequality is the comparison of the sum of the first n natural numbers, 1+2+...+n, to the sum of the first n squared natural numbers, 1^2+2^2+...+n^2. It can be shown that the latter sum is always greater than the former, which is a result of the summation inequality.

Are there any limitations to the summation inequality?

Yes, the summation inequality is limited to comparing the size of finite and infinite sums, and it can only be applied to certain types of series. It may not be applicable to more complex series, and it is important to carefully consider the conditions of the inequality before using it in a proof.

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