# A question about limit

1. Nov 30, 2013

### sunay

limn→∞($\frac{1}{n+1}$+$\frac{1}{n+2}$+...+$\frac{1}{4n}$)
Hi, when I first looked at this limit I thought that the solution is 0, but the assistant applied Riemann Sum and she found that this limit equals to ln4. Why this limit is not 0. I'm confused. Can you help me?

2. Nov 30, 2013

### Staff: Mentor

Why would you think that the limit is zero? This is a finite sum consisting of 3n terms, each of which is positive. You can't add a finite number of positive terms together and end up with zero.

I think you might be confused between the concepts of sequence and sum. A sequence is a list of numbers. In a sum such as the one you have, the numbers in a sequence are added together. The sequence {1/(n + 1), 1/(n + 2), ..., 1/(4n)} converges to 0 as n increases without bound, but the sum of the sequence converges to ln(4), as the teaching assistant showed.

3. Nov 30, 2013

### sunay

Actually I had seperated the limits and thought that each of them is 0, so the sum is 0. But now I understood I think. Thanks.

4. Nov 30, 2013

### mathwonk

you don't get the right result from adding the limits when the number of terms increases without bound as it does here. e.g. (1/n + 1/n + 1/n+....+ 1/n) with n terms, always equals 1, although each term individually approaches 0.

5. Nov 30, 2013

### lurflurf

The limit diverges in the usual sense. It is asymptotically equal to log(4n).

6. Nov 30, 2013

### Office_Shredder

Staff Emeritus
Those two statements sound contradictory to me.

7. Nov 30, 2013

### lurflurf

^Oops I had in mind limn→∞($\frac{1}{1}$+$\frac{1}{2}$+...+$\frac{1}{4n}$)

For that problem asked we have as stated a Riemann Sum
$$\lim_{n\rightarrow\infty}\sum_{k=1}^{3n}\frac{1}{n+k}=\lim_{n \rightarrow \infty}\frac{1}{n}\sum_{k=1}^{3n}\frac{1}{1+k/n}=\int_0^3 \frac{\mathrm{d}t}{1+t}=\log(4)$$

8. Nov 30, 2013

### lurflurf

The limit is not zero because
$$\lim_{n\rightarrow\infty}\sum_{k=1}^{3n}\frac{1}{n+k} \ne \sum_{k=1}^{3n}\lim_{n\rightarrow\infty}\frac{1}{n+k}=\sum_{k=1}^{3n} 0=0$$

We cannot move the limit inside the sum because the number of terms in the sum depends on the variable.