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I think it does but I'm having trouble showing it.

  • Using the integral test: the function $f(x)=\frac{1}{\ln(x^x+x^2)}$ is decreasing but I'm having trouble integrating $\int_{1}^{\infty} \frac{dx}{\ln(x^x+x^2)}$, trying the substitution $t=x^x$ didn't seem to work.

  • The root test gives $\sqrt[n]{\left |\frac{1}{\ln(n^n+n^2)}\right |} \to 1$ which doesn't imply anything.

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    You should be trying a limit comparison. First show that you can ignore the n^2. (The way to use the integral test is generally not to integrate the function itself, but to integrate obvious lower and/or upper bounds on the function which are easier to integrate.)2011-01-10

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Hint: Compare with $$\sum_{n=1}^{\infty} \frac{K}{n\log n}$$

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    If my calculations are correct then $K=1$ since: $$1 \leftarrow \frac{n\ln n}{\ln(2n^n)} \leq \frac{n\ln n}{\ln(n^n+n^2)} \leq \frac{n\ln n}{\ln(n^n)}=1$$ And since $\sum_{n=1}^{\infty} \frac{1}{n\log n}$ diverges, the mentioned series diverges.2011-01-10
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    @daniel: You only need one side of the comparison to prove divergence :-)2011-01-10
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    @Daniel: I was think more along the lines of $\frac{1}{\ln (n + n^2)} > \frac{1}{2(n \ln n)}$, but there are too many possibilities. What you have is correct too, in fact you have shown that any $0 < K \lt 1$ can be chosen for sufficiently large $n$.2011-01-10
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    @Moron: Yes but if I want to use the limit comparison then I need to show that $\frac{n\ln n}{\ln(n^n+n^2)} \to A \gt 0$, no?2011-01-10
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    @daniel: Yes if you want to use that. There are multiple ways :-) My comments were to show you that you can also use the integral test, as you intended to, originally.2011-01-10
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    @Moron: Ok, many thanks. One last question: my original intention was to find if $\sum \frac{\cos(2n)}{\ln(n^n+n^2)}$ converges absolutely. Since I thought $\sum \frac{1}{\ln(n^n+n^2)}$ converges I thought I have it easy. But now that it diverges I can look at $|\frac{\cos(2n)}{\ln(n^n+n^2)}|\geq \frac{\cos(2n)^2}{\ln(n^n+n^2)}$ and show the latter diverges, does that make sense?2011-01-10
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    @daniel: Yes that would prove it. Of course, I have no clue if that is true :-)2011-01-10