0

Determine whether the series converges or not: $$ \sum_{j=1}^{\infty} \left( e^{(-1)^{j}\sin(1/j)} - 1 \right) $$

My attempt:

\begin{align*} &\sum_{j=1}^{\infty} \left( e^{(-1)^{j}\sin(1/j)} - 1 \right) \\ &= \sum_{j=1}^{\infty} \left( \left( 1 + (-1)^{j}\sin(1/j)+\frac{(-1)^{2j}}{2!}\sin^2(1/j) + \dots \right) - 1 \right) \\ &= \sum_{j=1}^{\infty} \sum_{k=1}^{\infty} \frac{(-1)^{kj}}{k!} \sin^k(1/j). \end{align*}

By the M-test, $\sum_{k=1}^{\infty} \frac{(-1)^{kj}}{k!} \sin^k(1/j)$ converges uniformly. But I am not sure how do I conclude that $\sum_{j=1}^{\infty} \sum_{k=1}^{\infty} \frac{(-1)^{kj}}{k!} \sin^k(1/j)$ is convergent.

Anyone suggest me some idea.

Sangchul Lee
  • 181,930
User124356
  • 1,627

2 Answers2

1

Note that for all $j$ we have

$$-2<(-1)^j \sin(1/j)-1<0$$

Then

$$e^{(-1)^j \sin(1/j)-1}>e^{-2}$$

Thus

$$\sum_{j=1}^\infty e^{(-1)^j \sin(1/j)-1}>\sum_{j=1}^\infty e^{-2}=\infty$$

QC_QAOA
  • 12,277
1

Extending my previous comment, note that

$$ e^{(-1)^j\sin(1/j)} - 1 = \frac{(-1)^j}{j} + \mathcal{O}\left(\frac{1}{j^2}\right) \qquad\text{as}\quad j\to\infty. $$

So it follows that

$$ \sum_{j=1}^{\infty} \bigl( e^{(-1)^j\sin(1/j)} - 1 \bigr) = \sum_{j=1}^{\infty} \frac{(-1)^j}{j} + \sum_{j=1}^{\infty} \mathcal{O}\biggl( \frac{1}{j^2} \biggr), $$

which converges.

Addendum. If you are not familiar to the Big-O notation, here is a more detailed proof: Write $f(x) = e^{\sin x}$. Then by the Taylor's theorem, for each $x$ there exists $\xi$ between $0$ and $x$ such that

$$ f(x) = f(0) + f'(0)x + \frac{f''(\xi)}{2!}x^2 = 1 + x + \frac{f''(\xi)}{2}x^2. $$

Now if we specialize in $x \in [-1, 1]$, then $\xi$ also lies in $[-1, 1]$, and so,

$$ \left| f(x) - (1 + x) \right| \leq Mx^2, $$

where $M = \frac{1}{2}\max_{-1 \leq \xi \leq 1} |f''(\xi)|$. Then by plugging $x = (-1)^j/j$ for $j \geq 1$, which obviously lies in $[-1, 1]$, we get

$$ \left| e^{(-1)^j\sin(1/j)} - 1 - \frac{(-1)^j}{j} \right| \leq \frac{M}{j^2}. $$

So by the comparison test with $\sum_{j=1}^{\infty} \frac{1}{j^2}$,

$$ \sum_{j=1}^{\infty} \biggl( e^{(-1)^j\sin(1/j)} - 1 - \frac{(-1)^j}{j} \biggr) $$

converges absolutely. Since we know that $\sum_{j=1}^{\infty} \frac{(-1)^j}{j}$ also converges, adding these two series together proves that the original sum also converges.

Sangchul Lee
  • 181,930