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The question is to estimate the error for the series $$\sum_{n\to1}^\infty\frac{1}{n^2+1}$$ using the Remainder Estimate for the Integral Test: $$\int_{n+1}^\infty f(x)\,dx\le R_n\le\int_n^\infty f(x)\,dx$$ I did some hand calculations to determine that $n\ge 2000$ shows that the remainder $R_n\le0.0005$

I then tried:

s = Sum[1/(n^2 + 1), {n, 1, \[Infinity]}]
Assuming[n > 1, 
 Reduce[s - Sum[1/(1 + k^2), {k, 1, n}] < 0.0005, n, Integers]]

But that did not work. Does anyone have good suggestions how to use Mathematica to determine the remainder of this series?

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    $\begingroup$ not sure I understand exactly what the question is, but you need ~2000 terms to have r<0.0005 ( I just did Table[s-Sum..{n}] ) $\endgroup$ – george2079 Feb 27 '17 at 19:58
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    $\begingroup$ Are you looking for an expression that bounds the tail for any n? The Euler Mclaurin summation formula is one cute way to do it. $\endgroup$ – bobbym Feb 27 '17 at 20:12
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The questions say to use the integral remainder estimate.

tailUpper[n_] := Integrate[1/(x^2 + 1), {x, n, ∞}]
Reduce[tailUpper[n] <= Rationalize[0.0005], n, Integers]
(*  n ∈ Integers && n >= 2000  *)

Now get bounds on the error estimate:

tailLower[n_] := Integrate[1/(x^2 + 1), {x, n + 1, ∞}]
{tailLower[2000], tailUpper[2000]} // N

(*  {0.00049975, 0.0005}  *)
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  • $\begingroup$ Nice answer. This will be quite easy for my students to understand. Much appreciated. $\endgroup$ – David Feb 28 '17 at 1:37

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