# To check whether given series is convergent or divergent

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.

• Apply root test. It may help. – hemu Nov 18 '17 at 21:13
• But when ratio test fails,root test fails too. – Believer Nov 18 '17 at 21:14
• No, when the root tests fails, the ratio test fails. Not the other way round. – Duncan Ramage Nov 18 '17 at 21:16
• this sum does not converge – Dr. Sonnhard Graubner Nov 18 '17 at 21:16
• @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 $\frac{1}{2}$. – Duncan Ramage Nov 18 '17 at 21:41

## 2 Answers

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).

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

• 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 – adfriedman Nov 18 '17 at 21:38
• 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 – 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.

• 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. – adfriedman Nov 29 '17 at 11:04