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$$U_n = \arcsin\left(\frac{n+1}{2n+1}\right)-\arcsin\left(\frac{n-1}{2n-1}\right)$$

I have to determine if this series( $\sum U_n$ ) is converging or not.

I’ve tried to know if $\sum \sin(Un)$ is converging and given that $ \sin( U_n) \sim U_n $ and $\forall n \in \mathbb{N}$ $U_n \geq 0 $ to know if it converges but it didn't work out.

Thanks in advance!

dfnu
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abc
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2 Answers2

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Note that $$\lim_{n \to \infty} \frac{\arcsin\left(\frac{n+1}{2n+1}\right)-\arcsin\left(\frac{n-1}{2n-1}\right)}{1/n} = \frac{1}{\sqrt 3}$$ Since harmonic series diverges, by limit comparison test, so does $\sum \left(\arcsin\left(\frac{n+1}{2n+1}\right)-\arcsin\left(\frac{n-1}{2n-1}\right)\right)$.

NB: To justify this test we also need to show that the terms are positive. But I left this to you. ( You may use monotonicity of $\arcsin$, for example).

Azlif
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    We can’t use $\arcsin x\sim x$ in this case since x is not small. – user Nov 11 '19 at 12:52
  • In that way is really uncorrect, we should use $$\arcsin\left(\frac{n+1}{2n+1}\right)=\arcsin\left(\frac12+\frac{1}{2(2n+1)}\right)$$ and then expand. – user Nov 11 '19 at 13:11
  • @user how about that? I deleted the addendum – Azlif Nov 11 '19 at 13:15
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We can use that

$$\arcsin x + \arcsin y = \arcsin( x\sqrt{1-y^2} + y\sqrt{1-x^2}) $$

to prove that

$$ \arcsin\left(\frac{n+1}{2n+1}\right)-\arcsin\left(\frac{n-1}{2n-1}\right)\sim \frac1{\sqrt 3 n}$$

the series diverges by limit comparison test $\sum \frac1n$.

Refer also to

user
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