What is the lim of the sequence $(e^1 -1 ) + (e^{1/2} -1 ) + \dots + (e^{1/n} -1)$ as n approaches infinity ? Personally i got the answer $e-1$
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Can you please add your approach? It would be very helpful. – Umesh Shankar Feb 09 '23 at 14:09
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I did notice that this can be done using geomtric sequence sum as (a1(1-q^n))/1-q – ahmad Feb 09 '23 at 14:11
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How would i approach it other than using the geometric sum , – ahmad Feb 09 '23 at 14:14
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As the linked duplicate explains, use the definition of $e$ and compare your sum to the harmonic sum. – JMoravitz Feb 09 '23 at 14:22
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Its not clear what the final lim is – ahmad Feb 09 '23 at 14:36
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4You said that you got the answer $e-1$, but note that the very first term is $e-1$, and the following terms are all positive. – Théophile Feb 09 '23 at 14:50
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How is it not clear what the final limit is? The linked thread says it repeatedly. Do you not know what "diverges" means? I will have to say that this series seems a bit much to ask of a Calc I class, at least without copious hints. – Paul Sinclair Feb 10 '23 at 12:21