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$\sum_{n=0}^{\infty} \frac{n^n}{3^{1+3n}}$

I have, by Cauchy Criterion

$\lim_{n \rightarrow \infty} \sqrt[n]{|a_n|}= \frac{1}{27} \lim_{n \rightarrow \infty} \frac{n}{3^{1/n}}= ?$

How a finish?


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Note that the $n^{th}$ terms is $a_n = \dfrac13 \left(\dfrac{n}{27} \right)^n$.

For $n > 54$, we have that $a_n > \dfrac{2^n}3$.

Hence, what is $\lim_{n \to \infty} a_n$? What can you say about the series after finding what $\lim_{n \to \infty} a_n$ is?

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We're on a roll, man. – Pedro Tamaroff Nov 6 '12 at 23:55
@PeterTamaroff lol. One of those many days where my research work has taken back seat :-) – user17762 Nov 6 '12 at 23:56

Maybe you can tidy things up first

$$\sum\limits_{n = 0}^\infty {\frac{{{n^n}}}{{{3^{1 + 3n}}}}} = \frac{1}{3}\sum\limits_{n = 0}^\infty {\frac{{{n^n}}}{{{3^{3n}}}}} = \frac{1}{3}\sum\limits_{n = 0}^\infty {\frac{{{n^n}}}{{{27^n}}}} = \frac{1}{3}\sum\limits_{n = 0}^\infty {{{\left( {\frac{n}{27}} \right)}^n}} $$

Cauchy then says $$\lim \; a_n^{1/n}=\lim\; \frac n {27}\to\infty$$

So the sequence isn't summable. Maybe we can shed some light on this.

The sequence of the form $$a_n=\left(\frac{n}{\lambda}\right)^n$$

has the particular problem that $a_n\not \to0$. Indeed, Whenever $n>2\lambda $,


which means that $a_n\to \infty$.

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If all is about convergence or divergence of the series, you can also use the D'Alembert criterion (ratio test): $$ \lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} = \lim_{n \rightarrow \infty} \frac{\frac{(n+1)^{n+1}}{3\cdot 3^{3(n+1)}}}{\frac{n^{n}}{3\cdot 3^{3n}}} = \lim_{n \rightarrow \infty} \frac{(n+1)^{n+1}\cdot 3\cdot 3^{3n}}{n^{n} \cdot 3\cdot 3^{3(n+1)}} = \lim_{n \rightarrow \infty} \frac{(n+1)}{3^3} \left(1+\frac{1}{n}\right)^{n}$$ You should know that: $$\lim_{n \rightarrow \infty} \left(1+\frac{1}{n}\right)^{n}=e,$$ so: $$\lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} = \frac{e}{27}\lim_{n \rightarrow \infty}(n+1) = \infty. $$

Thus, the series diverges.

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