If Σ(n+1)! / 3^n converges
2007-06-07 22:31:47 · 2 answers · asked by headless_cross 2 in Science & Mathematics ➔ Mathematics
no
(n+1)!/3^n -> infinity as n -> infinity => not cvgt,
compare by a term by term expansion or use ratio test
an+1/an = n+1/3 -> infinity > 1
2007-06-07 22:42:54 · answer #1 · answered by Anonymous · 1⤊ 0⤋
Notice that (n+1)! grows much faster than 3^n
eg n=1 means (n+1)!=2 and 3^n=3
n=1 means (n+1)!=2 and 3^n=3
n=2 means (n+1)!=6 and 3^n=9
n=3 means (n+1)!=24 and 3^n=27
n=4 means (n+1)!=120 and 3^n=81
n=5 means (n+1)!=720 and 3^n=243
because the bottom gets multiplied by 3 while the top gets multiplied by successively larger integers.
Therefore the sum of the series will go to infinity as n goes to infinity.
Therefore the series diverges (does not converge)
2007-06-08 07:19:37 · answer #2 · answered by blind_chameleon 5 · 0⤊ 0⤋