Question

Lt (-1)^n does not exist but what is the
n->infty case with Lt (-1)^(n+1)
n->infty

Answer 1

Exactly the same.

Answer 2

...because both alternate without end +1, -1, +1, -1, ...

Answer 3

but how can u say the ans is zero in my previous question

Answer 4

because it was all over (2n-1) and the limit as n --> infty of
1/(2n-1)
is zero

Answer 5

You should write out the first twenty terms or so of your series
a_n = ( (-1)^n + 1 ) / ( 2n - 1 )
so you get a sense of of it. I think it's intuitively obvious that this does have limit zero.

a1 = 0
a2 = 2/3
a3 = 0
a4 = 2/7
a5 = 0
a6 = 2/11
a7 = 0
a8 = 2/15
a9 = 0
a10 = 2/19
....