# Finding 1/n + 1/(n+1) + + 1/2n in terms of the Euler's constant

1. Sep 12, 2009

### lkh1986

1. The problem statement, all variables and given/known data

By using the Euler's constant (0.577...), express the limit of the sum 1/n + 1/(n+1) + ... + 1/2n explicitly as n goes to infinity.

2. Relevant equations

The limit as n goes to infinity for [1/1 + 1/2 + ... + 1/n - In (n)] is the Euler constant, 0.577...

It has been deduced and shown previously that the limit of the sum 1/n + 1/(n+1) + ... + 1/2n is between 0.5 and 1. And the value lies somewhere around 0.6-0.75.

3. The attempt at a solution

I think the word "explicitly" in the question means that we need to express the answer in terms of the Euler's constant.

limit as n goes to infinity for [1/1 + 1/2 + ... + 1/n - In (n)] = 0.577...

Let a (subscript k) = 1/k + 1/(k+1) + ... + 1/2k and k<n

limit as k goes to infinity for [1/1 + 1/2 + ...1/(k-1) + 1/k + 1/(k+1)+...+1/2k +1/(2k+1) ... 1/n - In (n)] = 0.577...

limit as k goes to infinity for [1/1 + 1/2 + ...1/(k-1) + a(subscript k) + 1/(2k+1) ... 1/n - In (n)] = 0.577...

limit as k goes to infinity for [a(subscript k)] = 0.577... - limit as k goes to infinity [1/1 + 1/2 + ...1/(k-1)] - limit as k goes to infinity [(1/(2k+1) + ... + 1/n] + limit as k goes to infinity [In(n)]

Then I don't know how to continue. Perhaps I need to define other variables? It seems that k and n are a bit confusing...

Thanks.

2. Sep 13, 2009

### VietDao29

This is wrong!! What is n in your expression? Is n related somehow to k?

------------------------------

Ok, here's how I would tackle the problem. Let's denote the Euler Constant to be $$\gamma$$.

Let $$x_n = \frac{1}{1} + \frac{1}{2} + ... + \frac{1}{n} - \ln(n)$$

We know that:

$$\gamma = \lim_{n \rightarrow \infty} x_n = \lim_{n \rightarrow \infty} \left( \frac{1}{1} + \frac{1}{2} + ... + \frac{1}{n} - \ln (n) \right)$$

Since the limit of the sequence xn exists, all of its sub-sequences should have the same limit, too, right? Now consider the sub-sequence x2n

$$\lim_{n \rightarrow \infty} x_{2n} = \lim_{n \rightarrow \infty} \left( \frac{1}{1} + \frac{1}{2} + ... + \frac{1}{2n} - \ln (2n) \right) = \gamma$$

$$\Rightarrow \lim_{n \rightarrow \infty} x_{2n} = \lim_{n \rightarrow \infty} \left[ {\color{red}\left( \frac{1}{1} + \frac{1}{2} + ... + \frac{1}{n} \right)} + {\color{blue}\left( \frac{1}{n + 1} + ... + \frac{1}{2n} \right)} - \ln (2n) \right] = \gamma$$

You want to find the limit of the blue part, right?

Now, look at the red one, do you realize anything? Does it look "nearly" the same as something that you've already known? What can you do to make use of the Euler's Constant?

Well, let's see if you can find a way to continue. :)

3. Sep 13, 2009

### lkh1986

Oh, the subsequence concept, yup. That's useful.

So, gamma = [lim as n goes to infinity (1/1+1/2+...+1/n)] + [lim as n goes to infinity (1/(n+1) + ... + 1/2n)] - [lim as n goes to infinity (In (2n))]

Gamma = Gamma + lim as n goes to infinity (In (2n)) +[lim as n goes to infinity (1/(n+1) + ... + 1/2n)] - [lim as n goes to infinity (In (2n))]

Canceling gamma at both sides, we have:
[lim as n goes to infinity (1/(n+1) + ... + 1/2n)] = [lim as n goes to infinity (In (2n))] - lim as n goes to infinity (In (2n))

[lim as n goes to infinity (1/(n+1) + ... + 1/2n)] = lim as n goes to infinity [In (2n/n)] = lim as n goes to infinity In (2) = In(2) = 0.693...

I used my Texas Instrument to calculate the value. When n = 71, the value obtained is 0.703723..., which is very close to In(2), so I believe In 2 is the correct answer?

Thanks for the help. ;)

4. Sep 13, 2009

### VietDao29

Well, you're very close to correct..

Remember that, you can only break the limit of 2 sums into the sum of 2 limit, when both of the 2 limits exist:

$$\lim (x_n + y_n) = \lim x_n + \lim y_n$$ only holds if the limits of both sequences xn, and yn exist.

Both expressions diverge, so you cannot split it like that. try this instead:

$$\Rightarrow \lim_{n \rightarrow \infty} x_{2n} = \lim_{n \rightarrow \infty} \left[ {\color{red}\left( \frac{1}{1} + \frac{1}{2} + ... + \frac{1}{n} -\ln (n) \right)} + \left( {\color{blue}\left( \frac{1}{n + 1} + ... + \frac{1}{2n} \right)} - \ln (2n) + \ln(n) \right) \right] = \gamma$$

5. Sep 13, 2009

### lkh1986

Hm, everything looks good, but I think In 2 is the limit of 1/(n+1) + ... + 1/2n. The question asks for 1/n + 1/(n+1) + ... + 1/2n, so the 1/n term is missing there. ;( But In 2 is correct I think.

6. Sep 13, 2009

### VietDao29

You can modify it a little bit. It's still the same. How about, putting the term 1/n into the blue parentheses, and modifying the ln(...) terms a little bit? :)

7. Sep 13, 2009

### D H

Staff Emeritus
$$\Rightarrow \lim_{n \rightarrow \infty} x_{2n} = \lim_{n \rightarrow \infty} \left[ {\color{red}\left( \frac{1}{1} + \frac{1}{2} + ... + \frac{1}{n-1} -\ln (n-1) \right)} + \left( {\color{blue}\left( \frac{1}{n} + \frac{1}{n + 1} + ... + \frac{1}{2n} \right)} - \ln (2n) + \ln(n-1) \right) \right] = \gamma$$

8. Sep 13, 2009

### lkh1986

Yay, this is great.
So, lim [1/1 + 1/2 + ... + 1/(n-1) - In(n-1)] = gamma

So the equation can be reduced to lim (1/n + 1/(n+1) + ... + 1/(2n)) = lim In[2n/(n-1)] = lim In 2 = In 2. Is this correct?

Thanks both of you for the help. ;)

9. Sep 13, 2009

### VietDao29

Looks fine to me. :)

10. Sep 13, 2009

### lkh1986

I am so happy to finally solve the problem. ;) Thanks again and have a nice day!