Solving Limits of Sums: Ideas Needed

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The discussion centers on the limit of the sum lim_{x → ∞} ∑_{j=0}^x e^{-j/x}. Initial thoughts suggest that the jth term approaches 1, indicating divergence, but further analysis reveals that the expression actually converges, as confirmed by the integral test. Participants clarify that the inequality ∑_{i=0}^N e^{-i/N} ≥ (N + 1)(1/e) holds true for this specific sum, due to the presence of N + 1 terms, each being at least 1/e. The conversation emphasizes the importance of rigorous mathematical reasoning in determining convergence versus divergence. Overall, the expression converges despite initial assumptions of divergence.
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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?
 
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It looks to me that the jth term -> 1 for any j, which makes the sum divergent to ∞.
 
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.
 
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.
 
Apteronotus said:
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.
Apteronotus said:
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.
 
Yes, I meant to say diverges.
Thanks again for the thorough explanation.
 
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