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Determine whether the series converges or diverges (integral test) $$\sum_{n=1}^\infty \frac{1}{2^ {\ln n}}$$

Unsure how to proceed with this problem, the hint I was shown is that you need to check it with the integral test but I'm unsure how to integrate this.

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    $\begingroup$ If you like, then use the integral test to test the convergence or divergence of $\sum 1/n^p$ as the answer stated below. $\endgroup$ – xbh Dec 15 '18 at 2:07
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Before we try to apply any tests, we should try to put that general term in a more convenient form. An exponential raised to a logarithm? That should simplify. $$\frac1{2^{\ln n}} = \frac1{\left(e^{\ln 2}\right)^{\ln n}} = \frac1{\left(e^{\ln n}\right)^{\ln 2}} = \frac1{n^{\ln 2}}$$ Now that it's a fixed power of $n$, can you see what to do?

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  • $\begingroup$ I can see this diverges by test for p-series. So that is one solution to the answer. Thanks. $\endgroup$ – Luke D Dec 15 '18 at 1:57
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Use the Cauchy Condensation Test for $f(x) = \dfrac{1}{2^{\ln x}}\implies \displaystyle \sum_{n=1}^\infty 2^nf(2^n)=\displaystyle \sum_{n=1}^\infty2^n\cdot\dfrac{1}{2^{n\ln2}}= \displaystyle \sum_{n=1}^\infty (2^{(1-\ln 2)})^n=\infty$ ,since $2^{(1-\ln 2)}> 1$. Thus the given series diverges to $\infty$ as well.

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$\int_1^{\infty}2^{-\ln x}\operatorname dx=[\frac{x2^{-\ln x}}{1-\ln2}]_1^{\infty}=[\frac{xe^{-\ln x\ln2}}{1-\ln2}]_1^{\infty}=\lim_{x\to\infty}\frac {x\cdot(\frac1x)^{\ln2}-1}{1-\ln2}=\infty$.

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