Proof of Divergence of a Series

  • Thread starter Arkuski
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  • #1
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Prove that the series [itex]\displaystyle\sum_{k=1}^{\infty}\sqrt[k]{k+1}-1[/itex] diverges.

I thought that I could show the [itex]n^{th}[/itex] term was greater than [itex]\frac{1}{n}[/itex] but this is turning out to be more difficult than I imagined. Is there a neat proof that [itex]n^n>(n+1)^{n-1}[/itex]?
 

Answers and Replies

  • #2
Dick
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##(k+1)^\frac{1}{k}=\exp(\log((k+1)^\frac{1}{k}))##. Simplify the log a little and think about what the series expansion of ##e^x## looks like.
 

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