# Limits of sums

1. Feb 5, 2010

### Apteronotus

I'm stuck on how to approach the following problem.

$$lim_{x \rightarrow \infty} \sum_{j=0} ^x e^{-j/x}$$

Does anyone have any ideas?

2. Feb 5, 2010

### mathman

It looks to me that the jth term -> 1 for any j, which makes the sum divergent to ∞.

3. Feb 5, 2010

### Staff: Mentor

Looks like infinity to me. Expanding your sum gives
$$e^0 + e^{-1/N} +e^{-2/N} + ... + e^{-N/N}$$
In this expression there are N + 1 terms, the smallest of which is 1/e.

So,
$$\sum_{i = 0}^N e^{-i/N} \geq (N + 1)(1/e)$$

As N gets larger, so does (N + 1)/e.

4. Feb 5, 2010

### Apteronotus

Thanks you both for your replies. You are right, the expression does converge, as the integral test seems to varify.

Mark, I have one quick question. Could you explain why
$$\sum_{i = 0}^N e^{-i/N} \geq (N + 1)(1/e)$$

Is this a general statement or does it only apply in this case.

Thank you again.

5. Feb 5, 2010

### Staff: Mentor

No, what mathman and I are saying is that the expression diverges.
This statement applies to your sum. For each value of N, there are N + 1 terms being added. The smallest of these terms is 1/e, so we know that the sum has to be at least (N + 1) times 1/e, which is (N + 1)/e.

6. Feb 6, 2010

### Apteronotus

Yes, I meant to say diverges.
Thanks again for the thorough explanation.