Calculus I: infinite series

**NonCommAlg** · March 27th 2009, 07:51 PM

For what values of $0 < p \leq 1$ the alternating series $\sum_{n =1}^{\infty}(-1)^{n-1} \frac{\tan^{-1} n}{n^p}$ is convergent?

**bkarpuz** · March 28th 2009, 09:01 AM

Originally Posted by NonCommAlg

For what values of $0 < p \leq 1$ the alternating series $\sum_{n =1}^{\infty}(-1)^{n-1} \frac{\tan^{-1} n}{n^p}$ is convergent?

Answer. It is convergent for all $p\in(0,1]$ .
Solution. Due to Leibniz rule, it suffices to prove that $f(t,p):=\frac{\mathrm{ArcTan(t)}}{t^{p}}$ is decreasing for all $t\gg1$ and all $p\in(0,1]$ since $\lim\nolimits_{t\to\infty}f(t,p)=0$ for any $p\in(0,1]$ .
Therefore,
$\frac{\partial f}{\partial t}=\frac{1}{t^{p}}\bigg(\frac{1}{1+t^{2}}-p\frac{\mathrm{ArcTan(t)}}{t}\bigg)$
${\color{white}{\frac{\partial f}{\partial t}}}\leq\frac{1}{t^{p+1}}\bigg(\frac{1}{t}-p\mathrm{ArcTan(t)}\bigg).$
It is clear that the latter term in the above resulting is eventually negative since $1/t\to0$ and $p\mathrm{ArcTan(t)}\to p\pi/2$ .
Hence $f$ is decreasing for all $t$ sufficiently large and all $p$ positive, and this implies that the series is convergent for all $p\in(0,\infty)$ .

**NonCommAlg** · March 28th 2009, 11:35 AM

Originally Posted by bkarpuz

Answer. It is convergent for all $p\in(0,1]$ .
Solution. Due to Leibniz rule, it suffices to prove that $f(t,p):=\frac{\mathrm{ArcTan(t)}}{t^{p}}$ is decreasing for all $t\gg1$ and all $p\in(0,1]$ since $\lim\nolimits_{t\to\infty}f(t,p)=0$ for any $p\in(0,1]$ .
Therefore,
$\frac{\partial f}{\partial t}=\frac{1}{t^{p}}\bigg(\frac{1}{1+t^{2}}-p\frac{\mathrm{ArcTan(t)}}{t}\bigg)$
${\color{white}{\frac{\partial f}{\partial t}}}\leq\frac{1}{t^{p+1}}\bigg(\frac{1}{t}-p\mathrm{ArcTan(t)}\bigg).$
It is clear that the latter term in the above resulting is eventually negative since $1/t\to0$ and $p\mathrm{ArcTan(t)}\to p\pi/2$ .
Hence $f$ is decreasing for all $t$ sufficiently large and all $p$ positive, and this implies that the series is convergent for all $p\in(0,\infty)$ .

correct!

using limit to prove that a function is eventually negative was the idea of this problem! many calculus I students probably don't know this important point.

you didn't need the inequality though because you could have also argued that:

$\frac{\partial f}{\partial t}=\frac{p}{t^{p+1}} \left(\frac{t}{p(1+t^2)} - \tan^{-1}t \right) < 0,$ for $t \gg 0,$ because $\lim_{t \to\infty} \left(\frac{t}{p(1+t^2)} - \tan^{-1}t \right)= \frac{-\pi}{2} < 0.$

the reason that i assumed $0 < p \leq 1$ is that the series is obviously "absolutely" convergent for $p > 1.$

**bkarpuz** · March 28th 2009, 12:12 PM

Originally Posted by NonCommAlg

correct!

using limit to prove that a function is eventually negative was the idea of this problem! many calculus I students probably don't know this important point.

you didn't need the inequality though because you could have also argued that:

$\frac{\partial f}{\partial t}=\frac{p}{t^{p+1}} \left(\frac{t}{p(1+t^2)} - \tan^{-1}t \right) < 0,$ for $t \gg 0,$ because $\lim_{t \to\infty} \left(\frac{t}{p(1+t^2)} - \tan^{-1}t \right)= \frac{-\pi}{2} < 0$ .

the reason that i assumed $0 < p \leq 1$ is that the series is obviously "absolutely" convergent for $p>1$ .

A different solution can be given by using comparison.
It is obvious that
$\lim_{n\to\infty}\frac{\frac{\mathrm{ArcTan(n)}}{n ^{p}}}{\frac{1}{n^{p}}}=\frac{\pi}{2},$
which shows that $\frac{\mathrm{ArcTan(n)}}{n^{p}}$ is asymptotic to $\frac{\pi}{2n^{p}}$ .
And since the limiting series $\frac{\pi}{2}\sum\nolimits_{n\in\mathbb{N}}\frac{(-1)^{n-1}}{n^{p}}$ converges, the first one also converges.

**NonCommAlg** · March 28th 2009, 12:18 PM

Originally Posted by bkarpuz

A different solution can be given by using comparison.
It is obvious that
$\lim_{n\to\infty}\frac{\frac{\mathrm{ArcTan(n)}}{n ^{p}}}{\frac{1}{n^{p}}}=\frac{\pi}{2},$
which shows that $\frac{\mathrm{ArcTan(n)}}{n^{p}}$ is asymptotic to $\frac{\pi}{2n^{p}}$ .
And since the limiting series $\frac{\pi}{2}\sum\nolimits_{n\in\mathbb{N}}\frac{(-1)^{n-1}}{n^{p}}$ converges, the first one also converges.

well, i think we can't do that because limit comparison test is only applied to series with positive terms!

**bkarpuz** · March 28th 2009, 12:29 PM

Originally Posted by NonCommAlg

well, i think we can't do that because limit comparison test is only applied to series with positive terms!

I was now thinking about it!
But i guess a comparison can be given in this way,
if the series formed by the minorants of positive terms and majorants of negative terms converges, then the first one also converges.
For instance, rewrite the series in the following form
$\sum_{n\in\mathbb{N}}(-1)^{n}a_{n}=\sum\nolimits_{n\in\mathbb{N}}a_{2n}-\sum\nolimits_{n\in\mathbb{N}}a_{2n+1},$
then find two positive sequences $\{b_{n}\}_{n\in\mathbb{N}},\{c_{n}\}_{n\in\mathbb{ N}}$ such that $a_{2n}\geq b_{n}$ and $a_{2n+1}\leq c_{n}$ for all $n$ sufficiently large, and $\sum\nolimits_{n\in\mathbb{N}}(-1)^{n}d_{n}:=\sum\nolimits_{n\in\mathbb{N}}b_{n}-\sum\nolimits_{n\in\mathbb{N}}c_{n}$ converges.

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