Finding the set of all positive values of $a$ for which the series $$ \sum_{n=1}^\infty\left(\frac1n-\tan^{-1} \frac1n\right)^a $$ converges. How will it depend on the the value of a that is its power of the term? After expanding the arc tan term I get the form of summation of $[ -1/n^3(1/3+1/n^2+......]^a $. now how does it depend on a ?
problem on convergence of series $\sum_{n=1}^\infty\left(\frac1n-\tan^{-1} \frac1n\right)^a$
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calculus
sequences-and-series
convergence
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0On this website you're generally expected to write out your questions rather than inserting a picture, so that they can be found by search engines and hence more easily serve as future references – 2017-01-22
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1Sorry but I don't know how to type this symbols – 2017-01-22
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1I guess $\tan^{-1}$ is nothing else than $\arctan$ ... This said, if you know that $\arctan(t)\underset{0}{=}t-\frac{t^3}{3}+o(t^3)$, you are almost done. – 2017-01-22
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1Yes I got it. After solving the numerator should not have power less than1 otherwise it will diverge. It should be greater than equal to 1 so a will be greater than equal to 1/3. Am I right? – 2017-01-22
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0@shadow kh that's right. – 2017-01-22
2 Answers
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Hint. One may use a Taylor series expansion, as $n \to \infty$, $$ \arctan \frac1n=\frac1n+O\left(\frac1{n^3}\right) $$ giving, as $n \to \infty$, $$ \left(\frac1n-\arctan \frac1n\right)^a=O\left(\frac1{n^{3a}}\right) $$ I hope you can take it from here.
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Since you don't ask for which values of $a$ the series converges, here is an heuristic answer.
When $n$ goes to $\infty$, $\frac 1n$ goes to $0$ so $\tan^{—1}(1/n)$ goes to $0$.
So the general term (without the "$a$") of your series goes to $0$, so there is a reasonable chance that the series will converge.
It will converge if the general term goes to $0$ sufficiently fast.
That would be the case if $a$ is sufficiently big, but it won't be the case if $a$ is too small.