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I am just starting to learn Calculus. If anyone could help me that would be very useful. Thanks ahead

From here: how to prove $\sum {\frac{1}{n^{1+1/n}}}$ is divergent

I don't really get how to use induction from $\dfrac{1}{n ^ {1+ \frac{1}{n}}} \lt \dfrac{1}{2n}$

And another one is why $\sum_{n=1}^\infty {1\over {n ^ {1+ {1\over n}}}}$ is divergence and $\sum_{n=1}^\infty {1\over{n^{1+ {1\over \ln n} }}}$ Is not?

I mean $\sum_{n=1}^\infty {1\over{n^{1+ {1\over \ln (\ln n) } }}}$ (Sorry)

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  • $\begingroup$ From the last sentence of your post: what makes you think that $\sum_n \frac{1}{n^{1+\frac{1}{\ln n}}}$ is not divergent as well? It is divergent: you have $n^{\frac{1}{\ln n}} = e^{\frac{1}{\ln n}\cdot \ln n} = e^1 = e$, so $\frac{1}{n^{1+\frac{1}{\ln n}}}= \frac{e^{-1}}{n}$, and $\sum_n \frac{1}{n^{1+\frac{1}{\ln n}}}=\sum_n \frac{e^{-1}}{n} = \infty$. $\endgroup$ – Clement C. Feb 2 '16 at 3:55
  • $\begingroup$ Typesetting note: I do not recommend using \over as it can cause a great deal of confusion. Instead, \frac{ }{ } or if you want larger display, \dfrac{ }{ } are easier to manipulate. $\endgroup$ – JMoravitz Feb 2 '16 at 3:59
  • $\begingroup$ and why did $\sum_{n=0}^\infty {1\over{n^{1+ {1\over \ln {\ln (n)}} }}}$ converge? $\endgroup$ – Wakeme UpNow Feb 2 '16 at 4:05
  • $\begingroup$ Note that your series is undefined at the $n=0$ term, unless you interpret $1/\ln(0)$ as $0$. $\endgroup$ – alex.jordan Feb 2 '16 at 4:18
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After your edit: we will show that the series with general term $\frac{1}{n^{1+\frac{1}{\ln \ln n}}}$ (for $n\geq 3$) is convergent. To do so, we will compare it to the (convergent) series $\sum_{n=2}^\infty \frac{1}{n(\ln n)^2}$, which is a convergent Bertrand series.

Rewrite $$ n^{1+\frac{1}{\ln\ln n}} = n\cdot n^{\frac{1}{\ln\ln n}}= n\cdot e^{\frac{\ln n}{\ln\ln n}} $$ and $$n(\ln n)^2 = n\cdot e^{2\ln \ln n}$$

We have asymptotically that $\ln \ln n = o\left(\frac{\ln n}{\ln\ln n}\right)$, i.e. for $n$ big enough $\frac{\ln n}{\ln\ln n} > 2\ln \ln n$. This implies $$\frac{1}{n^{1+\frac{1}{\ln\ln n}}} < \frac{1}{n(\ln n)^2}$$ for $n$ big enough, and by comparison that the series $\sum_{n=3}^\infty \frac{1}{n^{1+\frac{1}{\ln \ln n}}}$ converges.

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HINT:

Note that we have

$$n^{1/\log(n)}=e$$


Note that in the original post, the series of interest was $\sum_{n=2}^{\infty}\frac{1}{n^{1+\frac{1}{\log(n)}}}$

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