[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]MarkFL said:Prove the following:
\(\displaystyle \sum_{k=n}^{2n-2}\frac{1}{k^2}<\frac{1}{2n-1}\) where \(\displaystyle 2\le n\)
Oops, seems like I was having a senior moment there. (Blush)Euge said:Hi Opalg,
At the end of your proof you write $\frac{1}{2n-2} < \frac{1}{2n-1}$, but in fact $\frac{1}{2n-2} > \frac{1}{2n-1}$.
Let's hope this attempt is better than the previous one.MarkFL said:Prove the following:
\(\displaystyle \sum_{k=n}^{2n-2}\frac{1}{k^2}<\frac{1}{2n-1}\) where \(\displaystyle 2\le n\)
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]
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]MarkFL said:... we want to add to the left:
\(\displaystyle -\frac{1}{n^2}+\left(\frac{1}{2(n+1)-2}\right)^2=\left(\frac{1}{2n}\right)^2-\frac{1}{n^2}=-\frac{3}{4n^2}\)
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]
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.
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.
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.
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.
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.