Convergence of the series $\sum_{n = 1}^{+\infty}{\left(n\sin{\frac{1}{n}}\right)^n}$

Question

I have to study the convergence of the series

$$ \sum_{n = 1}^{+\infty}{\left(n\sin{\frac{1}{n}}\right)^n} $$

and

$$ \sum_{n = 1}^{+\infty}{\left(\left(n\sin{\frac{1}{n}}\right)^n - 1\right)}. $$

I know I should study the limit

$$ \lim_{n\to +\infty}{\left(n\sin{\frac{1}{n}}\right)^n} $$

and that

$$ \lim_{n\to +\infty}{n\sin{\frac{1}{n}}} = 1 $$

but I don't see how it helps. Any ideas ?

Thank you in advance !

score 6 · Accepted Answer · answered Apr 13 '16 at 20:17

6

On the interval $(0,1)$ we have $$ 1-\frac{x^2}{6} \leq \frac{\sin x}{x}\leq e^{-x^2/6} $$ hence $\left(n\sin\frac{1}{n}\right)^n$ behaves like $e^{-\frac{1}{6n}}=1-\frac{1}{6n}+O\left(\frac{1}{n^2}\right)$ for large values of $n$ and the given series are divergent.

answered Apr 13 '16 at 20:17

Jack D'Aurizio

353,855

Thanks ! This helped me a lot. – Biblot Apr 15 '16 at 10:07

score 2 · Answer 2 · answered Apr 13 '16 at 20:40

2

It is easy to show from the mean-value theorem that $x-\frac16 x^3\le \sin(x)\le x$ for $0\le x\le 1$. Therefore,

$$n\sin(1/n)\ge 1 -\frac{1}{6n^2}$$

Using Bernoulli's Inequality, we find that

$$(n\sin(1/n))^n\ge \left(1 -\frac{1}{6n^2}\right)^n\ge 1-\frac{1}{6n}$$

Inasmuch as the general terms of the series do not approach $0$ as $n\to \infty$, the series diverges.

answered Apr 13 '16 at 20:40

Mark Viola

179,405

1

Since the second series diverges to $-\infty$, an upper bound is needed, no? – Vincenzo Oliva Apr 13 '16 at 20:44
@VincenzoOliva What do you mean by second series? – Mark Viola Apr 13 '16 at 20:50
$\sum ( n \sin(1/n) )^n-1$ goes to $-\infty$, so don't we need an upper bound on the general term? – Vincenzo Oliva Apr 13 '16 at 20:56
@VincenzoOliva Yes, for the second series, we need more than simply this lower bound. – Mark Viola Apr 13 '16 at 21:23

Vincenzo Oliva · Answer 3 · 2016-04-13T20:53:39.333

1

Since $\sin\frac1n=\frac1n-\frac1{6n^3}+O(1/n^5)$, it follows that $$\lim \left(n\sin\frac1n\right)^n=\left(1-\frac{1}{6n^2}\right)^n=1,$$whence the divergence of the first series. We now also know that the second series has indeed a tending-to-zero general term, but $$\left(n\sin\frac1n\right)^n-1<\left(1-\frac{1}{7n^2}\right)^n -1<-\frac{1}{8n}$$ and thus we have divergence to $-\infty$.

edited Apr 13 '16 at 20:53

answered Apr 13 '16 at 20:38

Vincenzo Oliva

7,249

(he used $\ln(1+x) \sim x$ when $x \to 0$ hence $\ln (1-\frac{a}{n^k})^n = n\ln (1-\frac{a}{n^k}) \sim -\frac{a}{n^{k-1}}$) – reuns Apr 13 '16 at 21:42

Convergence of the series $\sum_{n = 1}^{+\infty}{\left(n\sin{\frac{1}{n}}\right)^n}$

3 Answers3