How do you prove that #sum_(n=1)^oo (n^(1/n)-1)# diverges?

Question

Emma Aldridge · Accepted Answer

Use a comparison test against the harmonic series.

Use a comparison against the harmonic series: #sum_(n=1)^oo 1/n# which is known to diverge. Note that: #e^x = 1+x/(1!)+x^2/(2!)+x^3/(3!)+...# #n^(1/n) = e^(1/n ln n) = 1 + (1/n ln n)/(1!) + (1/n ln n)^2/(2!) + ...# When #n >= 3# we have #ln n > 1#, so: #n^(1/n) - 1 = (1/n ln n)/(1!) + (1/n ln n)^2/(2!) + ... > 1/n# Hence: #sum_(n=3)^oo (n^(1/n)-1) > sum_(n=3)^oo 1/n# diverges.

William Abdalla · Answer

undefined To prove that the series $ \sum_{n=1}^{\infty} \left( n^{\frac{1}{n}} - 1 \right) $ diverges, we can use the Limit Comparison Test. Let's consider the series $ \sum_{n=1}^{\infty} \frac{1}{n} $.

Now, we'll calculate the limit as $ n $ approaches infinity of the ratio of the terms of the given series to the terms of the series $ \sum_{n=1}^{\infty} \frac{1}{n} $:

$$ \lim_{n \to \infty} \frac{n^{\frac{1}{n}} - 1}{\frac{1}{n}} $$

By using L'Hôpital's Rule, we can rewrite this limit as:

$$ \lim_{n \to \infty} \frac{\frac{d}{dn}(n^{\frac{1}{n}} - 1)}{\frac{d}{dn}\left(\frac{1}{n}\right)} $$

Solving the derivatives:

$$ \lim_{n \to \infty} \frac{\frac{1}{n^2} \cdot n^{\frac{1}{n}} \cdot (1 - \ln(n))}{- \frac{1}{n^2}} $$

This simplifies to:

$$ \lim_{n \to \infty} n^{\frac{1}{n}} \cdot (1 - \ln(n)) $$

Now, as $ n $ approaches infinity, $ n^{\frac{1}{n}} $ approaches 1. Also, $ \ln(n) $ approaches infinity as $ n $ approaches infinity. Therefore, $ 1 - \ln(n) $ approaches negative infinity.

Since we have a product where one term approaches 1 and the other approaches negative infinity, the limit of the product is negative infinity.

By the Limit Comparison Test, since the limit is not finite and positive, the given series $ \sum_{n=1}^{\infty} \left( n^{\frac{1}{n}} - 1 \right) $ diverges.