Solving Limits of Sums: Ideas Needed

[Apteronotus]

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

<br /> lim_{x \rightarrow \infty} \sum_{j=0} ^x e^{-j/x}<br />

Does anyone have any ideas?

 

[mathman]

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

 

[Mark44]

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.

 

[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
<br /> \sum_{i = 0}^N e^{-i/N} \geq (N + 1)(1/e)<br />

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

Thank you again.

 

[Mark44]

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

No, what mathman and I are saying is that the expression diverges.

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

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

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.

 

[Apteronotus]

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