S= 1/2 + 1/4 + 1/6 + 1/8 +....+ 1/2n+....
S= (sum from n=1 to infinity) (e^(-n))/ (1+n^2)
I got that the first one diverges and the 2nd one converges, and used the ratio test on the first one and the integral and comparison tests on the 2nd. Is this correct? Thank you.
S= (sum from n=1 to infinity) (e^(-n))/ (1+n^2)
I got that the first one diverges and the 2nd one converges, and used the ratio test on the first one and the integral and comparison tests on the 2nd. Is this correct? Thank you.
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For the first, I use the comparison test (with the harmonic series, known to diverge), but you could also use the integral test. That's the usual way to show divergence in the harmonic series originally.
The ratio test should work on the second series e^(-(n+1)) / e^(-n) = 1/e, and (1/(n^2 + 1)) / (1/((n+1)^2 + 1) is (n^2 + 2n + 1)/(n^2 + 1) ... which tends to 1 as n->infinity. So the ratio of consecutive terms has a limit 1/e, and that limit is less than 1 in magnitude.
The ratio test should work on the second series e^(-(n+1)) / e^(-n) = 1/e, and (1/(n^2 + 1)) / (1/((n+1)^2 + 1) is (n^2 + 2n + 1)/(n^2 + 1) ... which tends to 1 as n->infinity. So the ratio of consecutive terms has a limit 1/e, and that limit is less than 1 in magnitude.