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Convergence of series $$\sum^{\infty}_{k=1}k^2\tan\frac{k+2}{k^2+5}$$

What I have tried: using $\tan x>x$ for $x>0$

$$\tan\frac{k+2}{k^2+5}>\frac{k+2}{k^2+5}$$

$$\sum^{\infty}_{k=1}k^2\tan\frac{k+2}{k^2+5}>\sum^{\infty}_{k=1}\frac{k^2(k+2)}{k^2+5}$$

jimjim
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jacky
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2 Answers2

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$\tan x \sim \sin x \sim x$ when $x \rightarrow 0$, where under "$\sim$" we understand limit $\dfrac{f}{g} \rightarrow 1$. So $$k^2\tan\frac{k+2}{k^2+5} \sim k^2 \dfrac{1}{k} = k$$

zkutch
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  • You should mention you're using equivalence of functions, and not the physicists ‘approximately equal to’, to be fully rigourous. – Bernard Jun 06 '20 at 00:33
  • Yes, I mean exactly the limit of $\frac{f}{g}$. I thought it's clear in calculus/math.analysis environment, but added anyway to answer. Thanks. As to be fully rigorous, we need note, that series is positive, which comes from $\frac{k+2}{k^2+5} \rightarrow 0+$. – zkutch Jun 06 '20 at 00:45
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A series $\sum a_n$ cannot converge unless $a_n \to 0$ as $ n \to \infty$. To show that $\sum \frac {k^{2}(k+2)} {k^{2}+5}$ is not convergent let us show that $\frac {k^{2}(k+2)} {k^{2}+5}$ does not tend to $0$. Dividing the numerator and denominator by $k^{2}$ you can write $\frac {k^{2}(k+2)} {k^{2}+5}=k\frac {\frac 2 k +1} {1+\frac 5 {k^{2}}}$. This tends to $(\infty) (1)=\infty$.

Kavi Rama Murthy
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