The problem asks us to determine if the series $\sum_{k=1}^{\infty} k \sin \frac{1}{k}$ converges or diverges.

AnalysisSeries ConvergenceLimit TestDivergence TestLimitsTrigonometric Functions
2025/3/7

1. Problem Description

The problem asks us to determine if the series k=1ksin1k\sum_{k=1}^{\infty} k \sin \frac{1}{k} converges or diverges.

2. Solution Steps

We will use the limit test for divergence. If limkak0\lim_{k \to \infty} a_k \neq 0, then the series ak\sum a_k diverges.
In our case, ak=ksin1ka_k = k \sin \frac{1}{k}.
We want to find limkksin1k\lim_{k \to \infty} k \sin \frac{1}{k}.
Let x=1kx = \frac{1}{k}. As kk \to \infty, x0x \to 0. Thus,
limkksin1k=limx0sinxx \lim_{k \to \infty} k \sin \frac{1}{k} = \lim_{x \to 0} \frac{\sin x}{x}
We know that limx0sinxx=1\lim_{x \to 0} \frac{\sin x}{x} = 1.
Thus,
limkksin1k=1 \lim_{k \to \infty} k \sin \frac{1}{k} = 1
Since limkksin1k=10\lim_{k \to \infty} k \sin \frac{1}{k} = 1 \neq 0, the series k=1ksin1k\sum_{k=1}^{\infty} k \sin \frac{1}{k} diverges by the divergence test.

3. Final Answer

Diverges

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