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Math Help - Series convergence question

  1. #1
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    Series convergence question

    I'm stuck on a series convergence/divergence problem. Does this coverge or diverge: the sum of [(the nth root of 2)] - 1 as n goes to infinity? I tried the ratio test, but I got 1 as the limit, so that doesn't tell me anything. What should I try next?

    Thanks.
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  2. #2
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    Quote Originally Posted by virtuoso735 View Post
    I'm stuck on a series convergence/divergence problem. Does this coverge or diverge: the sum of [(the nth root of 2)] - 1 as n goes to infinity? I tried the ratio test, but I got 1 as the limit, so that doesn't tell me anything. What should I try next?

    Thanks.
    Try a limit comparison with \sum_{n=1}^\infty \frac{1}{n} i.e. consider \lim_{n \to \infty} \frac{2^{1/n} - 1}{\frac{1}{n}}
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  3. #3
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    Hmm, I tried using the limit comparison test above and it came out to be 0, which is inconclusive is it not?
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    If we consider \lim_{n \to \infty} \frac{2^{1/n} - 1}{\frac{1}{n}}, if we let n = \frac{1}{x} then  \lim_{x \to 0} \frac{2^{x} - 1}{x} = \lim_{x \to 0} 2^{x} \ln 2 = \ln 2 (by L'Hopital's rule). Since  \sum_{n=1}^\infty \frac{1}{n} diverges, by the LCT, the original series diverges.
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  5. #5
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    You mean 0+

    Quote Originally Posted by danny arrigo View Post
    If we consider \lim_{n \to \infty} \frac{2^{1/n} - 1}{\frac{1}{n}}, if we let n = \frac{1}{x} then  \lim_{x \to 0^+} \frac{2^{x} - 1}{x} = \lim_{x \to 0^+} 2^{x} \ln 2 = \ln 2 (by L'Hopital's rule). Since  \sum_{n=1}^\infty \frac{1}{n} diverges, by the LCT, the original series diverges.
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  6. #6
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    Quote Originally Posted by matheagle View Post
    You mean 0+
    Of course - thanks for pointing that for the OP.
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