... diverges and You can prove that considering that is...
But a primitive of the function under integral sign in (2) is so that the integral in (2) diverges and the same is for the series (1)...
1st) I need to investigate convergence using convergence of integral feature
From limit I see that it's 0, but how do I prove it using that method?
2nd) Here I need to investigate convergence using series required condition of convergence
for the 1)st question you know that 1/ 1/n n N (natural numbers). and you know that is the harmonic series that diverges. so by comparison test diverges. since 1/n diverges c 0 c 1/n diverges. therefore the radius of convergence is 0.
2nd) there is a theorem that states that converges = 0
which means if 0 diverges. since 1 n 0 n N, for a given n you can find n N such that cos n 0, which means n . therefore diverges. once again the radius of convergence is 0.
hope this is helpful.