Does $\sum\limits_{n=0}^\infty \frac{e^n\sin n}{n}$ converge or diverge?

Question

The comparison test does not work on this, so I'm stuck trying to find a way to prove that it diverges. I know it definitely diverges. Any solutions?

Answer 1

Consider the points $p_n=(\cos n,\sin n),$ $ n=1,2,\dots $ on the unit circle. The $p_n$ march around the circle infinitely many times in steps of of arc length $1.$ Let $A$ denote the arc on the unit circle between the points $(1/\sqrt 2,1/\sqrt 2)$ and $(-1/\sqrt 2,1/\sqrt 2).$ The length of this arc is $\pi/2>1.$ It is impossible for our sequence $p_n$ to jump over $A,$ as the $p_n$ are marching around in steps of arc length $1.$ Thus $p_n\in A$ for infinitely many $n.$ For each such $n,$ we have $\sin n \ge 1/\sqrt 2.$ It follows that $(e^n\sin n)/n \ge e^n/(\sqrt 2\cdot n)$ for infinitely many $n.$ Thus the terms of our series do not approach $0,$ which shows the series diverges.

Answer 2

$$ \lim_{n \rightarrow \infty} \frac{e^n\sin n}{n} \neq 0 $$ So the series diverges using the nth term test for divergence.

Answer 3

Take the root test. We have $\limsup_{n\to\infty}(|e^n\sin(n)|)^{1/n} = e> 1$ so it diverges.

Answer 4

It suffices to show that $\sin(n)$ fails to approach $0$, which means that your series fails to converge by the $n$th term test (that is, the sequence of summands fails to approach $0$).

Some proofs of that $\sin(n)$ fails to converge to $0$ are given here.

I would suspect, however, that if you are learning about testing for convergence/divergence of series for the first time, you would not be expected to prove such a result. If it comes up, I would state without proof that $\sin(n)$ does not have a limit as $n\to \infty$, just as the function $f(x) = \sin(x)$ fails to have a limit as $x \to \infty$.

Answer 5

The series does not converge because the general term does not tend to zero. To prove this, it is enough to show that $\sin n$ does not converge. Here is an elementary way to do it: suppose that $\sin n\to x\in \mathbb R$.

Then, of course,

$\sin 2n\to x$ and $\sin (2(n+1))\to x.$

And so

$\sin n\to x\Rightarrow \cos 2n\to 1-2x^2\Rightarrow \cos (2(n+1))\to 1-2x^2$

But now,

$\sin 2=\sin(2(n + 1) - 2n)=\sin(2(n + 1))\cdot\cos 2n-\sin 2n\cdot \cos (2(n+1))\to 0 $, which is absurd.

Answer 6

If $|\sin(n)\,|\le\sin\left(\frac12\right)$, then $|\cos(n)\,|\ge\cos\left(\frac12\right)$ and $$ \begin{align} |\sin(n+1)\,| &=|\cos(n)\sin(1)+\sin(n)\cos(1)\,|\\ &\ge\cos\left(\tfrac12\right)\sin(1)-\sin\left(\tfrac12\right)\cos(1)\\ &=\sin\left(\tfrac12\right) \end{align} $$ This means that either $|\sin(n)\,|\ge\sin\left(\frac12\right)$ or $|\sin(n+1)\,|\ge\sin\left(\frac12\right)$.

Since $e^n\gt1+n$, we know that $\frac{e^n}n\gt1$.

Therefore, either $\left|\frac{e^n\sin(n)}n\right|\ge\sin\left(\frac12\right)$ or $\left|\frac{e^{n+1}\sin(n+1)}{n+1}\right|\ge\sin\left(\frac12\right)$. That is, the terms do not tend to $0$.

Answer 7

Assuming $\frac{e^n \sin n}{n}\to 0$ as $n\to +\infty$ we have $\left|\sin(n)\right|\leq \frac{n}{2e^n}\leq\sin\frac{1}{2^n}$ for any $n$ sufficiently large, hence $\left|\cos n\right|\geq \cos\frac{1}{2^n}$. On the other hand such assumptions lead to $$ \left|\sin(n+1)\right|\geq \left|\left|\sin(n)\right|\cos(1)-\left|\cos(n)\right|\sin(1)\right|\geq \frac{4}{5} $$ for any $n$ sufficiently large, contradiction.