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Test for convergence:

Alexmahone

Active member
Jan 26, 2012
268
Test for convergence: $\sum_1\sin\left(\frac{1}{n}\right)$

My attempt:

$\sin\left(\frac{1}{n}\right)\sim\frac{1}{n}$

Since $\sum_1\frac{1}{n}$ diverges, $\sum_1\sin\left(\frac{1}{n}\right)$ also diverges by the asymptotic convergence test.

Is that correct?
 
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Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
Test for convergence: $\sum_1\sin\left(\frac{1}{n}\right)$

My attempt:

$\sin\left(\frac{1}{n}\right)\sim\frac{1}{n}$

Since $\sum_1\frac{1}{n}$ diverges, $\sum_1\sin\left(\frac{1}{n}\right)$ also diverges by the asymptotic convergence test.

Is that correct?
What you did looks ok to me. You could also use the limit comparison test with $a_n=\sin\left(\frac{1}{n}\right)$ and $b_n=\frac{1}{n}$. Since $\lim \dfrac{\sin\left(\frac{1}{n}\right)}{\frac{1}{n}}=1$, $\sum a_n$ and $\sum b_n$ have the same behavior; thus $\sum \sin\left(\frac{1}{n}\right)$ diverges since the harmonic series diverges.

I hope this helps!
 

Alexmahone

Active member
Jan 26, 2012
268
What you did looks ok to me. You could also use the limit comparison test with $a_n=\sin\left(\frac{1}{n}\right)$ and $b_n=\frac{1}{n}$. Since $\lim \dfrac{\sin\left(\frac{1}{n}\right)}{\frac{1}{n}}=1$, $\sum a_n$ and $\sum b_n$ have the same behavior; thus $\sum \sin\left(\frac{1}{n}\right)$ diverges since the harmonic series diverges.

I hope this helps!
Thanks!
 
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