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.

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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

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).

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

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.