Using Taylor expansion check if series : ∑n=1∞n(arctan⁡(n+1)−arctan⁡n) converges.

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Using Taylor expansion check if series :  ∑n=1∞n(arctan⁡(n+1)−arctan⁡n)  converges.

Marcus Herman · Accepted Answer

Before we apply Taylor series, we will manipulate a bit. Note thattan⁡(arctan⁡(n+1)−arctan⁡n)=11+n(n+1)Here we have used the formula for tan⁡(u−v)Thus arctan⁡(n+1)−arctan⁡n=arctan⁡(11+n(n+1))Now the power series for arctan⁡x will do the job. For the range we are interested in, it is an alternating series. In particular, we conclude that our difference of arctan&#039;s is less than 11+n(n+1), which is less than 1n2.

Mary Nicholson · Accepted Answer

Hint: Note that the Taylor series of arctan⁡(x+1)−arctan⁡(x) at x=∞ is  1x2−1x3+1x5+O(1x6)  ⇒arctan⁡(n+1)−arctan⁡(n)∼1n2, n→∞

user_27qwe · Accepted Answer

Hint We have arctanx+arctan⁡1x=π/2 so n(arctan⁡(n+1)−arctan⁡n)=n(arctan⁡1n−arctan⁡1n+1)∼1n3/2 hence we can conclude the convergence of the series by comparaison with the Riemann series.

Using Taylor expansion check if series : \sum_{n=1}^\infty\sqrt{n}(\arctan(n+1)-\arctan n) converges.

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