Expert Answer to "Prove that the series ∑i=1∞1nsin⁡(1n) is divergent"

High School
Geometry
Trigonometry
Trigonometric equations and identities

Ernest Ryland

Answered question

2022-01-17

Prove that the series

\sum_{i = 1}^{\infty} \frac{1}{\sqrt{n}} \sin (\frac{1}{\sqrt{n}})

is divergent

Answer & Explanation

Navreaiw

Beginner2022-01-18Added 34 answers

For

0 \leq x \leq \frac{π}{2}

, we have

\frac{2}{π} x \leq \sin (x) \leq x

Therefore use

\frac{2}{π} x \leq \sin (x), x = \frac{1}{\sqrt{n}}

, and get after multiplication

\frac{2}{π} \frac{1}{n} \leq \sin (\frac{1}{\sqrt{n}})

Paul Mitchell

Beginner2022-01-19Added 40 answers

A variant is to use $lim_{u \to 0} \frac{\sin u}{u} = 1$ and in particular this term is $> \frac{1}{2} for | u | ≪ 1$ small enough.
Now write $\frac{1}{\sqrt{n}} \sin (\frac{1}{\sqrt{n}}) = \frac{1}{n} \times \frac{\sin (\frac{1}{\sqrt{n}})}{\frac{1}{\sqrt{n}}} > \frac{1}{n} \times \frac{12}{}$ for n large enough.
And still conclude by comparison with $\sum \frac{1}{n}$
Note: a faster way is to use equivalents, the limit 1 means $\frac{1}{\sqrt{n}} \sin (\frac{1}{\sqrt{n}}) \sim \frac{1}{n}$

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