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$$\sum_{n=2}^\infty\frac{\ln^{5}(2n^{7}+13)+10\sin(n)}{n\, \ln^{6}(n^{\frac{7}{8}}+2\sqrt{n}-1)\ln(\ln(n+(-1)^{n}))}$$

Check convergence.

What I've done before:

$$\ln^{5}(2n^{7}+13)+10\sin(n)>\ln^{5}n\Leftarrow\frac{\ln^{5}(2n^{7}+13)}{\ln^{5}(n)}>10\Leftarrow\frac{7^{5}\ln^{5}(n)}{\ln^{5}(n)}>10\Leftarrow7^{5}>10$$

$$\, \ln^{6}(n^{\frac{7}{8}}+2\sqrt{n}-1)\ln(\ln(n+(-1)^{n}))<\ln^{6}(n)\cdot \ln(\ln(n))\Leftrightarrow$$

$$\frac{\ln^{6}(n^{\frac{15}{16}})}{\ln^{6}(n)}<\frac{\ln(\ln(n))}{\ln(\ln(n+(-1)^{n})}\Leftrightarrow\frac{15}{16}<\frac{\ln(\ln(n))}{\ln(\ln(n+(-1)^{n})}$$

Series $$\sum_{n=2}^\infty\frac{1}{n\cdot\ln\, n\cdot\ln\, \ln\, n}$$ don't converge

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1 Answer 1

I only do the test non-rigorously, but I think it is enough.

The term $10\sin(n)$ can be ignored. Similarly the term $n+(-1)^{n}$ is just $n$. Put everything together and ignore nonsense terms we need to work out the convergence of $$\sum^{\infty}_{n=2}|\frac{\ln^{5}(2n^{7})}{n\ln^{6}(n^{7/8})\ln\ln(n)}$$

It is clear that $\ln(2n^{7})=\ln2+7\ln(n)$, so the top term above is approximately $7^{5}\ln^{5}(n)$. The top term below is $n*(7/8)^{6}\ln^{6}(n)\ln\ln(n)$. So the real comparision is $\sum \frac{1}{n\ln(n)\ln\ln(n)}$'s convergence. By comparison test we need to test $\int \frac{1}{x\ln(x)\ln\ln(x)}$. This function is $\ln\ln\ln(x)$'s derivative, hence diverges. So the original series must diverge.

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Could You do it without integrals? We can't use them. –  Joggi Jan 9 '13 at 14:15
    
Sure, check Gauss's test:mathworld.wolfram.com/GausssTest.html –  Bombyx mori Jan 9 '13 at 14:18
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