I have to prove the uniform convergence of this sequence $f_n(x)=\tan^{-1}nx$ in $[a,b],a>0$
What I have reached so far:
$$|f_n(x)-f(x)|=\left|\tan^{-1}nx-\frac\pi 2\right|=\tan^{-1}nx-\frac\pi 2<\epsilon$$
How do I proceed further ?
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@HagenvonEitzen I have edited the question to include the interval – Aman Mittal Jul 20 '13 at 12:29
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@AmanMittal Alright, it would even work with $[a,\infty)$ provided $a>0$. – Hagen von Eitzen Jul 20 '13 at 12:32
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Recalling the Taylor series of $\arctan(t)$ at $t=\infty$
$$ \frac{\pi}{2} -{\frac {1}{t}}+O \left( {t}^{-3} \right) $$
$$ \implies \arctan( nx )=\frac{\pi}{2}-{\frac {1}{ny}}+O \left( {n}^{-3} \right) $$
$$ \implies \arctan( nx )-\frac{\pi}{2}= -\frac{1}{nx}+O \left( {n}^{-3} \right) $$
$$ \implies \arctan( nx )-\frac{\pi}{2} \sim -\frac{1}{nx}. $$
Now, you can advance to finish the problem.
Mhenni Benghorbal
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Nice!! so $m$ may be chosen as any value greater than $-1\over \epsilon x$ – Aman Mittal Jul 20 '13 at 12:51
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Hint: $\lim_{x\to+\infty}\tan^{-1} x=\frac\pi 2$ means that for every $\epsilon>0$ there exists $x_0$ such that $x>x_0$ implies $\left|\tan^{-1}x-\frac\pi2\right|<\epsilon$.
Hagen von Eitzen
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1I have already reached that far (as indicated). The problem is what should i do after this – Aman Mittal Jul 20 '13 at 12:33
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