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

[MarkFL]

Prove the following:

$$\sum_{k=n}^{2n-2}\frac{1}{k^2}<\frac{1}{2n-1}$$ where $$2\le n$$

 

[Opalg]

Prove the following:

$$\sum_{k=n}^{2n-2}\frac{1}{k^2}<\frac{1}{2n-1}$$ where $$2\le n$$

[sp]$$\sum_{k=n}^{2n-2}\frac{1}{k^2}$$ 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]

 

[Euge]

Hi Opalg,

Spoiler

At the end of your proof you write $\frac{1}{2n-2} < \frac{1}{2n-1}$, but in fact $\frac{1}{2n-2} > \frac{1}{2n-1}$.

 

[Opalg]

Hi Opalg,

Spoiler

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)

 

[Opalg]

Prove the following:

$$\sum_{k=n}^{2n-2}\frac{1}{k^2}<\frac{1}{2n-1}$$ where $$2\le n$$

Let's hope this attempt is better than the previous one.
[sp]$k^2>k^2-1 = (k-1)(k+1)$. Therefore $$\frac1{k^2} < \frac1{(k-1)(k+1)} = \frac1{2(k-1)} - \frac1{2(k+1)},$$ 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) $$\frac1{2n-1}$$, 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]

 

[Euge]

Hi Opalg,

Spoiler

Yes, this proof works! Nice one! (Cool)

 

[MarkFL]

Let's hope this attempt is better than the previous one.
[sp]$k^2>k^2-1 = (k-1)(k+1)$. Therefore $$\frac1{k^2} < \frac1{(k-1)(k+1)} = \frac1{2(k-1)} - \frac1{2(k+1)},$$ 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) $$\frac1{2n-1}$$, 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:

Spoiler

We are given to prove:

$$\sum_{k=n}^{2n-2}\frac{1}{k^2}<\frac{1}{2n-1}$$ where $$2\le n$$

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

$$\sum_{k=2}^{2\cdot2-2}\frac{1}{k^2}<\frac{1}{2\cdot2-1}$$

$$\sum_{k=2}^{2}\frac{1}{k^2}<\frac{1}{2\cdot2-1}$$

$$\frac{1}{2^2}<\frac{1}{2^2-1}$$

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:

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

And we want to add to the right:

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

So, what we need to show is that:

$$-\frac{3}{4n^2}<-\frac{2}{4n^2-1}$$

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

$$8n^2<12n^2-3$$

Subtract through by $8n^2$:

$$0<4n^2-3$$

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

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

to $P_n$:

$$\sum_{k=n}^{2n-2}\frac{1}{k^2}<\frac{1}{2n-1}$$

to get:

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

which then simplifies to:

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

We have derived $P_{n+1}$ from $P_n$ thereby completing the proof by induction.

 

[Opalg]

[sp]

... we want to add to the left:

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

I think that what you need to add on the left is $$-\frac1{n^2} + \frac1{(2n-1)^2} + \frac1{(2n)^2}$$ (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]

 

[MarkFL]

[sp]
I think that what you need to add on the left is $$-\frac1{n^2} + \frac1{(2n-1)^2} + \frac1{(2n)^2}$$ (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]

Spoiler

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. :)

 

[MarkFL]

Spoiler

Because we find:

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

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

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

Hence:

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

$$\sum_{k=n}^{2n-2}\frac{1}{k^2}<\frac{1}{n-\frac{1}{2}}-\frac{1}{2n-\dfrac{3}{2}}$$

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

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

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