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The given series is $\sum_{n=1}^{\infty}\left(\frac{1}{n^{1+1/n}}\right)$.I try ratio test but it gives 1.Any hint.

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    $\begingroup$ Apply root test. It may help. $\endgroup$ – hemu Nov 18 '17 at 21:13
  • $\begingroup$ But when ratio test fails,root test fails too. $\endgroup$ – Believer Nov 18 '17 at 21:14
  • $\begingroup$ No, when the root tests fails, the ratio test fails. Not the other way round. $\endgroup$ – Duncan Ramage Nov 18 '17 at 21:16
  • $\begingroup$ this sum does not converge $\endgroup$ – Dr. Sonnhard Graubner Nov 18 '17 at 21:16
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    $\begingroup$ @omkarGirkar Consider the sequence given by $1/2, 1, 1/8, 1/4, 1/32, 1/16, \dots"$, that is, $a_n = \frac{1}{2^{n + 1}}$ if $n$ is even, and $a_{n} = \frac{1}{2^{n - 1}}$ if $n$ is odd. The limit for the ratio test doesn't even exist, but the root test confirms it converges with a limit of $\frac{1}{2}$. $\endgroup$ – Duncan Ramage Nov 18 '17 at 21:41
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As $$\lim_{n\to\infty} \frac{1}{n^{1/n}} = 1$$ it follows that

$$ \frac{1}{n^{1/n}} > \frac{1}{2}$$ for all $n \geq N$ for some fixed $N$ (in fact, $1/n^{1/n}$ hits a minimum larger than $1/ 2$ at $n=3$ ($0.69\!\dotsc$), so this holds for any $N\geq1$, but that isn't actually necessary to know).

$$\sum_{n=1}^{\infty} \frac{1}{n^{1+1/n}} = \sum_{n=1}^{N-1} \frac{1}{n^{1+1/n}} + \sum_{n=N}^{\infty} \frac{1}{n^{1/n}} \cdot \frac{1}{n^1} > \sum_{n=1}^{N-1} \frac{1}{n^{1+1/n}} + \frac{1}{2}\underbrace{\sum_{n=N}^{\infty}\frac{1}{n}}_{\to\infty}$$

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  • $\begingroup$ Only kidding... I had meant to say it monotonically decreases before hitting a minimum at $\frac{1}{e^{1/e}}$ and then monotonically increases, but I clearly got bored of typing $\endgroup$ – adfriedman Nov 18 '17 at 21:38
  • $\begingroup$ Ok..I understand your answer but just now I come to know one result which will really make the solution even more easy..can I post my answer $\endgroup$ – Believer Nov 18 '17 at 21:52
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If $a(n)$$>$$0$ and $\lim na(n)$ is not $0$ as $n$ goes to $\infty$,then $\sum a(n)$ is divergent.So in this case $\lim na(n)$=$1$,So easily we can say that $\sum a(n)$ is divergent.

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  • $\begingroup$ This is just an application of the limit comparison test. This arguably needs more machinery to properly prove than the answer I posted, but it works. $\endgroup$ – adfriedman Nov 29 '17 at 11:04

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