Prove that 1/(n+1) + 1/(n+2) + ... + 1/(2n) > 13/24

[songoku]

Homework Statement

Prove that for any naturam number n > 1 :
$\frac{1}{n+1} + \frac{1}{n+2} + \frac{1}{n+3} + ... + \frac{1}{2n} > \frac{13}{24}$

Homework Equations

Not sure

The Attempt at a Solution

$\frac{1}{n+1} + \frac{1}{n+2} + \frac{1}{n+3} + ... + \frac{1}{2n} > \frac{1}{2n} + \frac{1}{2n} + \frac{1}{2n} + ... + \frac{1}{2n} = \frac{n}{2n} = \frac{1}{2}$

Then I stuck...

Thanks

 

[Fightfish]

I'm not sure what approach your instructor had in mind in setting the problem, but unless I'm mistaken, we can actually prove a stronger inequality bound.

If we define
$$f(n) = \frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n},$$
then it is relatively straightforward to show that $f(n+1) > f(n) \,\, \forall \,\, n$, and so it must be the case that $f(n) \geq f(2)$.

It is also possible to prove that $\lim_{n \to \infty} f(n) = \ln 2$ as an interesting exercise.

 

[Mark44]

Homework Statement

Prove that for any naturam number n > 1 :
$\frac{1}{n+1} + \frac{1}{n+2} + \frac{1}{n+3} + ... + \frac{1}{2n} > \frac{13}{24}$

Homework Equations

Not sure

The Attempt at a Solution

$\frac{1}{n+1} + \frac{1}{n+2} + \frac{1}{n+3} + ... + \frac{1}{2n} > \frac{1}{2n} + \frac{1}{2n} + \frac{1}{2n} + ... + \frac{1}{2n} = \frac{n}{2n} = \frac{1}{2}$

Then I stuck...

Thanks

You can do this by mathematical induction, which might be the strategy your instructor had in mind.
Proving a base case with n = 2 or n = 3 is straightforward.
Then, assume that the proposition is true for n = k; i.e., that $\frac 1 {k + 1} + \frac 1 {k + 2} + \dots + \frac 1 {2k} > \frac {13} {24}$ (the induction hypothesis).
Finally, use the induction hypothesis to show that $\frac 1 {k + 2} + \frac 1 {k + 3} + \dots + \frac 1 {2k} + \frac 1 {2k + 1} + \frac 1 {2k + 2} > \frac {13} {24}$ must be true, as well.

 

[songoku]

You can do this by mathematical induction, which might be the strategy your instructor had in mind.
Proving a base case with n = 2 or n = 3 is straightforward.
Then, assume that the proposition is true for n = k; i.e., that $\frac 1 {k + 1} + \frac 1 {k + 2} + \dots + \frac 1 {2k} > \frac {13} {24}$ (the induction hypothesis).
Finally, use the induction hypothesis to show that $\frac 1 {k + 2} + \frac 1 {k + 3} + \dots + \frac 1 {2k} + \frac 1 {2k + 1} + \frac 1 {2k + 2} > \frac {13} {24}$ must be true, as well.

Let me try:
(i) For n = 2
1/3 + 1/4 = 7/12 = 14/24 > 13/24 so it is true for n = 2

(ii) Assume it is true for n = k
$\frac 1 {k + 1} + \frac 1 {k + 2} + \dots + \frac 1 {2k} > \frac {13} {24}$

(iii) For n = k + 1
$\frac 1 {k + 2} + \frac 1 {k + 3} + \dots + \frac 1 {2k} + \frac 1 {2k + 1} + \frac 1 {2k + 2}$

$= \frac 1 {k + 1} + \frac 1 {k + 2} + \frac 1 {k + 3} + \dots + \frac 1 {2k} + \frac 1 {2k + 1} + \frac 1 {2k + 2} - \frac 1 {k + 1}$

$> \frac {13} {24} + \frac 1 {(2k + 1) (2k + 2)} > \frac {13} {24}$

Is this correct?

 

[haruspex]

Let me try:
(i) For n = 2
1/3 + 1/4 = 7/12 = 14/24 > 13/24 so it is true for n = 2

(ii) Assume it is true for n = k
$\frac 1 {k + 1} + \frac 1 {k + 2} + \dots + \frac 1 {2k} > \frac {13} {24}$

(iii) For n = k + 1
$\frac 1 {k + 2} + \frac 1 {k + 3} + \dots + \frac 1 {2k} + \frac 1 {2k + 1} + \frac 1 {2k + 2}$

$= \frac 1 {k + 1} + \frac 1 {k + 2} + \frac 1 {k + 3} + \dots + \frac 1 {2k} + \frac 1 {2k + 1} + \frac 1 {2k + 2} - \frac 1 {k + 1}$

$> \frac {13} {24} + \frac 1 {(2k + 1) (2k + 2)} > \frac {13} {24}$

Is this correct?

Looks good.

 

[songoku]

Sorry for late reply

Thank you very much