$$\int^\infty_1\frac{1}{x^{1+1/x}}dx$$
I'm preparing for exams. I would also like to know what are some commonly used methods to show an improper integral diverges?
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3Look at $\lim\limits_{x\to\infty} x^{1/x}$. – Daniel Fischer Apr 02 '14 at 21:42
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@DanielFischer: good hint. Actually, all we really need is that $x^{1/x}\le e^{1/e}$. – robjohn Apr 02 '14 at 22:08
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True, @robjohn. But here, I think the limit is easier to find than the bound if one does not already know the bound. – Daniel Fischer Apr 02 '14 at 22:12
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@DanielFischer: that's true. I did say it was a good hint. I was just noting this because it would not require us to say, "for some $x$ large enough..." – robjohn Apr 02 '14 at 22:18
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$$ \begin{align} \int_n^\infty\frac1{x^{1+1/x}}\,\mathrm{d}x &\ge\int_n^\infty\frac1{x^{1+1/n}}\,\mathrm{d}x\\ &=nn^{-1/n} \end{align} $$ and $$ \begin{align} \lim_{n\to\infty}nn^{-1/n} &=\lim_{n\to\infty}n\lim_{n\to\infty}n^{-1/n}\\ &=\infty\cdot1 \end{align} $$
Another approach is to notice that $x^{1/x}\le e^{1/e}$.
robjohn
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We can expand the denominator of the integrand
\begin{align} &x^{1+1/x}=x x^{1/x}=x\exp \left( \frac{1}{x} \ln x \right)\\ =& x \cdot \left(1 + \frac{1}{x} \ln x + \frac{1}{2x^2} (\ln x)^2 + \cdots \right)\\ =& x+\ln x + \frac{1}{2x} (\ln x)^2 + O\left( \frac{(\ln x)^3}{x^2} \right) \\ =& O(x). \end{align}
Therefore, the improper integral diverges.
robit
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Hint: Consider the series $$\sum_{n=1}^{\infty}\frac{1}{n^{1+1/n}}$$
and apply the limit comparison test and the integral test.
PVAL-inactive
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2@user130916: the ratio of the terms tends to $1$, so the ratio test is inconclusive. – robjohn Apr 02 '14 at 22:12
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