Expert Answer to "Finding limn→∞∑r=1ncot−1(n2(r2+r+1)+2n+1n2+n)"

kuntungw3

Answered question

2022-01-26

Finding

lim_{n \to \infty} \sum_{r = 1}^{n} \cot^{- 1} (\frac{n^{2} (r^{2} + r + 1) + 2 n + 1}{n^{2} + n})

Allison Compton

Beginner2022-01-27Added 16 answers

Doing some basic manipulation we get the following terms

\cot^{- 1} (\frac{n (r^{2} + r)}{n + 1} + \frac{n + 1}{n})

. Now I let

\frac{n}{n + 1} = a

so we have

\cot^{- 1} (a (r^{2} + r) + \frac{1}{a}) = \arctan (\frac{a}{a^{2} (r^{2} + r) + 1})

we have y=ar+a,x=ar thus it becomes

\arctan (a r + a) - \arctan (a r)

thus sum is

\sum_{r = 1}^{\infty} \arctan (\frac{n}{n + 1} (r + 1)) - \arctan (\frac{n r}{n + 1}) = \frac{π}{2} - \frac{π}{4} = \frac{π}{4}

Howard Gallagher

Beginner2022-01-28Added 13 answers

\cot^{- 1} (\frac{n^{2} (r^{2} + r + 1) + 2 n + 1}{n^{2} + n}) \to \frac{π}{4} as n \to \infty

for each r so the required limit is

\infty

. [ Look at the sum from r=1 to N for fixed N and then let

N \to \infty

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