determine if the series converges or diverges:
1/(ksqrt(k^2+1)) from k = 1 to infinity
5 answers
as k→∞, √(k^2+1) ≈ k, so the terms approach 1/k^2 so it converges.
would you use the limit comparison test?
that would be good, since √(k^2+1) > √k^2 = k
im just confused on how to show my work with the limit comparison test. would my b_n = 1/sqrt(k^2) and then my a_n would be 1/(ksqrt(k^2+1))
and since b_n converges a_n will also converge?
and since b_n converges a_n will also converge?
yes, though you want b_n = 1/k^2
then a_n = 1/(k√(k^2+1)) < 1/(k√k^2) < 1/k^2 which converges.
then a_n = 1/(k√(k^2+1)) < 1/(k√k^2) < 1/k^2 which converges.