is that the direct comp test?
the question was sumation n=1 oo 2^n / n^20. i used the ratio test to get 2/n^20(n+1)^20.
on the sheet of paper i have it says 0<b_n<=a_n will diverge
wouldnt 1/n be b_n in this case?
1/n is bigger than 2/n^20(n+1)^20 so it wouldnt work...
edit:
oh wait...it = 0 so it converges....maybe i did something wrong then? because the series is supposed to diverge...
can you show me how to solve sumation n=1 oo 2^n / n^20 to show that it diverges using the ratio test?
Almost! I used the "squeeze theorem" in disguise.
Theorem: If {a_n} and {b_n} are convergent sequences to the same limit L and {c_n} is a sequence such that:
a_n <= c_n <= b_n for sufficiently large n
Then {c_n} is convergent to L as well.
Theorem: If {a_n} is a convergent sequence to 0 and {c_n} is a sequence such that:
|c_n|<=a_n for sufficiently large n
Then {c_n} is convergent to 0.
Proof: We can write |c_n|<=a_n differently:
-a_n<=c_n<=a_n
But the limit of a_n is 0 therefore the limit of -a_n is also 0.
Hence by above theorem the limit c_n exists and equals 0.