series $\sum_{n=1}^{\infty}\left(\frac{1}{3}\right)^n\cdot\left(\frac{n+1}{n}\right)^{n^2}$

Question

Series $$\sum_{n=1}^{\infty}\left(\frac{1}{3}\right)^n\cdot\left(\frac{n+1}{n}\right)^{n^2}$$

I tried Abel, Dirichlet theorems and it seems like divergent series but I don’t know the series a can compare to (for proving of divergence)

//i’m sorry that I can’t post images cuz due to reputation, this is my first question :>

Is your series $\sum_{n=1}^{\infty}\left(\frac{1}{3}\right)^n\cdot\left(\frac{n+1}{n}\right)^{n^2}$? — coreyman317, Jan 15 '20 at 14:18
since $\dfrac{e}{3} \lt 1$, I would expect this series to converge — Henry, Jan 15 '20 at 14:20
The sum of the series seems to be slightly less than $6.22513$ — Henry, Jan 15 '20 at 14:30

score 2 · Accepted Answer · answered Jan 15 '20 at 14:20

2

to find $$\dfrac13\cdot \lim_{n\to\infty}\left(1+\dfrac1n\right)^n=\dfrac e3<1$$

answered Jan 15 '20 at 14:20

lab bhattacharjee

Thanks so much, i forgot to put 1/n to 1/3 :D – gladozya Jan 15 '20 at 14:31
It is not necessary for the proof of convergence but $\lim\limits_{n\to\infty}\dfrac{\left(1+\frac1n\right)^{n^2}/3^n}{e^n/3^n} = \dfrac{1}{\sqrt{e}}$ – Henry Jan 15 '20 at 14:36

Peter Szilas · Answer 2 · 2020-01-15T14:55:31.153

0

$0 < (1/3)^n[(1+1/n)^n]^n <$

$(1/3)^ne^n=(e/3)^n=:a^n$, where $a<1$.

$\sum a^n$ is convergent, now use comparison test.

Used: $(1+1/n)^n$ is increasing, bounded from above by $e$.

Recall: $\lim_{n\rightarrow \infty}(1+1/n)^n=e.$

edited Jan 15 '20 at 14:55

answered Jan 15 '20 at 14:39

Peter Szilas

2 Answers2