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anonymous

  • 5 years ago

convergent or divergent? infinity summation (k/k+1) ^(k^2) k=1

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  1. anonymous
    • 5 years ago
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    so I went ahead and looked at what the sum is doing before it reaches the power and factored out a k\[\sum_{1}^{\infty}(k(1)/k(1+(1/k)))^(k^2) = \sum_{1}^{\infty}(1/(1+(1/k))^(k^2)\] if we take the limit as k goes to inf of that last form, before it goes to the infinite power, it converges to 1. (1/(1+0) ^inf.

  2. anonymous
    • 5 years ago
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    wonderful, thank you. so you used the root test correct? it dosen't matter if the function is ^k or ^(k^2)

  3. anonymous
    • 5 years ago
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    no, I just resolved the base of the sum and took the limit. Because it resolved to 1 there it doesn't really matter how how and fast and goes up in the power, it will always be 1. In general when going to the power of extraordinarily large numbers it's not really as relevent, but yu should still be able to intuit, that ^k^2 is going up faster than ^k.

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