test the convergency of $\arctan(k)$ series

Question

Test the convergence of series

$\sum^{\infty}_{k=1}(-1)^{k-1}\arctan(3k)$

What i have done is

As we know that $$\arctan(x)<\frac{\pi}{2}\forall x\in\mathbb{R}$$ $$\sum^{\infty}_{k=1}(-1)^{k-1}\arctan(3k)<\sum^{\infty}_{k=1}(-1)^{k-1}\frac{\pi}{2}.$$

After that how do i find whether the series is convergentcor divergent

Help me please

Your inequality for the sum is incorrect since for negative terms, the inequality for the inverse tangent is reversed. — Gary, Apr 03 '20 at 08:14

Qurultay · Accepted Answer · 2020-04-03T08:42:07.080

3

For the series $\sum_{k=1}^\infty (-1)^{k-1}\arctan(3k)$ to be convergent, first of all, we need to check if the general term of series tends to $0$ or not. In this case: $$\lim_{n\to\infty}(-1)^{k-1}\arctan(3k)=\frac{\pi}{2}\ne0 $$ thus, the series diverges.

edited Apr 03 '20 at 08:42

answered Apr 03 '20 at 08:16

Qurultay

5,224

ndid not get anything from here please explain me – jacky Apr 03 '20 at 08:21
if $\sum a_n$ converges, then $\lim_{n\to\infty}a_n=0$ – Qurultay Apr 03 '20 at 08:23
But in above linit $\displaystyle \lim_{n\rightarrow \infty}(-1)^{n-1}\cdot \arctan(3n)$=$\pm 1 \times \frac{\pi}{2}=\pm\frac{\pi}{2}$ is it right and can we say it is converge. – jacky Apr 03 '20 at 08:30
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@jacky No, since $\lim_{n\to\infty}(-1)^{k-1}\arctan(3k)\ne0$, therefore the first condition for a series to be convergent is not satisfied, therefore the series is divergent – Qurultay Apr 03 '20 at 08:34
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Hint :$ \lim_{k} \arctan (3k) \not =0$. Hence? – Peter Szilas Apr 03 '20 at 08:37
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@jacky Check the edit please. – Qurultay Apr 03 '20 at 08:42
Thanks so much Qurultay. – jacky Apr 03 '20 at 09:54

test the convergency of $\arctan(k)$ series

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