test the convergence of the series sigma tan inverse 1/n
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tan(x)=sin(x)cos(x)tan(x)=sin(x)cos(x).
For 0≤x≤π/20≤x≤π/2, sin(x)≤xsin(x)≤x. Integrating this from 00 to tt, ∫t0sin(x)dx≤∫t0xdx∫0tsin(x)dx≤∫0txdx or −cos(x)|t0≤x22|t0−cos(x)|0t≤x22|0t or 1−cos(t)≤t2/21−cos(t)≤t2/2 or cos(t)≥1−t2/2cos(t)≥1−t2/2.
Therefore tan(x)≤x1−x2/2tan(x)≤x1−x2/2 so that, for n≥1n≥1, tan(1/n)≤1/n1−(1/n)2/2<2/ntan(1/n)≤1/n1−(1/n)2/2<2/n so that tan2(1/n)<4/n2tan2(1/n)<4/n2 and the sum of these converges by any test you want to use.
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For 0≤x≤π/20≤x≤π/2, sin(x)≤xsin(x)≤x. Integrating this from 00 to tt, ∫t0sin(x)dx≤∫t0xdx∫0tsin(x)dx≤∫0txdx or −cos(x)|t0≤x22|t0−cos(x)|0t≤x22|0t or 1−cos(t)≤t2/21−cos(t)≤t2/2 or cos(t)≥1−t2/2cos(t)≥1−t2/2.
Therefore tan(x)≤x1−x2/2tan(x)≤x1−x2/2 so that, for n≥1n≥1, tan(1/n)≤1/n1−(1/n)2/2<2/ntan(1/n)≤1/n1−(1/n)2/2<2/n so that tan2(1/n)<4/n2tan2(1/n)<4/n2 and the sum of these converges by any test you want to use.
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