Proving S2n-Sn is greater than a half in terms of the harmonic series

In summary, the homework statement is that Sn is the sum of 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}...+\frac{1}{n}. If n is even, then Sn is the sum of half of these numbers, rounded down. If n is odd, Sn is the sum of the numbers from 1 to (n-1). If m is even, then S2m+2-Sm+1 is the sum of half of these numbers, rounded down. If m is odd, S2m+2-Sm is the sum of the numbers from 1 to (m-1).
  • #1
gottfried
119
0

Homework Statement


Let Sn = 1+[itex]\frac{1}{2}[/itex]+[itex]\frac{1}{3}[/itex]+[itex]\frac{1}{4}[/itex]...+[itex]\frac{1}{n}[/itex]. Show that |S2n-Sn|[itex]\geq[/itex] [itex]\frac{1}{2}[/itex]


Homework Equations





The Attempt at a Solution



So I'm going to try and use induction

Base case let n=1

|S2n-Sn| = [itex]\frac{1}{2}[/itex]

So true for base case

Assume true for case that n=m

So if n=m+1

|S2m+2-Sm+1|= |S2m+2-S2m+S2m-Sm+1|

Our induction assumption tells us that |S2m-S2m+2|[itex]\geq[/itex] [itex]\frac{1}{2}[/itex] and S2m+2-S2m = [itex]\frac{1}{2m+1}[/itex]+[itex]\frac{1}{2m+2}[/itex]> 0

Therefore |S2m+2-Sm|= |S2m+2-S2m+S2m-S2m+2|> [itex]\frac{1}{2}[/itex]

Have I used induction properly? Since I haven't used the base case at all the proof feels a little like I've made the assumption that it is true and then shown that it is true but at the same time I feel like I haven't broken any of the rules of induction. I'm I wrong?
 
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  • #2
gottfried said:

Homework Statement


Let Sn = 1+[itex]\frac{1}{2}[/itex]+[itex]\frac{1}{3}[/itex]+[itex]\frac{1}{4}[/itex]...+[itex]\frac{1}{n}[/itex]. Show that |S2n-Sn|[itex]\geq[/itex] [itex]\frac{1}{2}[/itex]


Homework Equations





The Attempt at a Solution



So I'm going to try and use induction

Base case let n=1

|S2n-Sn| = [itex]\frac{1}{2}[/itex]

So true for base case

Assume true for case that n=m

So if n=m+1

|S2m+2-Sm+1|= |S2m+2-S2m+S2m-Sm+1|

Our induction assumption tells us that |S2m-S2m+2|[itex]\geq[/itex] [itex]\frac{1}{2}[/itex] and S2m+2-S2m = [itex]\frac{1}{2m+1}[/itex]+[itex]\frac{1}{2m+2}[/itex]> 0
I don't quite follow that last step. But since ##\{S_n\}## is an increasing sequence you have$$
|(S_{2m+2}-S_{2m}) + (S_{2m}-S_{m+1})|=(S_{2m+2}-S_{2m}) + (S_{2m}-S_{m+1})\ge \frac 1 2$$since the quantities in parentheses are positive.

[Edit]: Ignore the highlighted paragraph. I wasn't completely awake. Look below:

But this problem hardly needs induction. Write out ##S_{2n}-S_n##, look at how many terms there are, and underestimate each by the smallest one.
 
Last edited:
  • #3
So there are n terms between S2n and Sn the smallest of which is [itex]\frac{1}{2n}[/itex]

It follows that |S2n - Sn|>n([itex]\frac{1}{2n}[/itex] ) =[itex]\frac{1}{2}[/itex]

That is frustratingly simple. I feel silly but thanks a lot.
 

1. What is the harmonic series?

The harmonic series is an infinite series in mathematics that represents the sum of the reciprocals of the natural numbers (1, 2, 3, ...). The series can be written as 1 + 1/2 + 1/3 + 1/4 + 1/5 + ... and it diverges, meaning it does not have a finite sum.

2. What is S2n-Sn?

S2n-Sn is the difference between two partial sums of the harmonic series. Specifically, it is the difference between the sum of the first 2n terms and the sum of the first n terms. In other words, it represents the sum of the terms from n+1 to 2n.

3. How do you prove that S2n-Sn is greater than a half?

To prove that S2n-Sn is greater than a half, we can use a mathematical induction proof. First, we can show that the statement is true for n=1 by substituting n=1 into the formula for S2n-Sn and showing that it is greater than 1/2. Then, we can assume the statement is true for n=k and use this assumption to prove that it is also true for n=k+1. This will show that the statement is true for all natural numbers and therefore, S2n-Sn is greater than a half.

4. Why is proving S2n-Sn greater than a half important?

Proving that S2n-Sn is greater than a half is important because it helps us understand the behavior of the harmonic series. The harmonic series is a divergent series, meaning it does not have a finite sum. However, by showing that S2n-Sn is greater than a half, we can see that the partial sums of the series are getting closer and closer to infinity. This understanding can have applications in other areas of mathematics and science.

5. Can this proof be extended to other series?

Yes, this proof can be extended to other series that have similar properties to the harmonic series. For example, it can be used to prove that the sum of the reciprocals of the square numbers (1, 1/4, 1/9, ...) also diverges. This proof technique can also be applied to other types of series, such as geometric series, to show whether they converge or diverge.

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