Question

Determine if the sequence is converging or diverging. If converging, find the limit.

a(n)= (1+(-1)^n)/n

Why is this converging?

Answer 1

please we have:
\[{a_n} = \frac{2}{n}\]
if n is even, and:
\[{a_n} = 0\]
if n is odd

Answer 2

I'm sorry. I still don't know why.

Answer 3

when n is an even number then, we have:
\[1 + {\left( { - 1} \right)^n} = 1 + 1 = 2\]
whereas when n is an odd number, we have:
\[1 + {\left( { - 1} \right)^n} = 1 - 1 = 0\]

Answer 4

Yes, doesnt that indicate that it isn't monotonic?

Answer 5

\[
a_n = \frac{1+(-1)^n}{n} = \frac 1n + \frac{(-1)^n}{n} \\
\text{As }n \rightarrow \infty, ~~~ \frac 1n \rightarrow 0 ~~~ \text {and }~~~
\frac{(-1)^n}{n} \rightarrow 0 \\
a_n \rightarrow 0 \text{ when } n \rightarrow \infty \\
\text{The series converges}.
\]

Answer 6

I meant the SEQUENCE converges. Not the series.

Answer 7

Thanks. Sorry but I don't understand why (-1)^n / n = 0 as n -> infinity. I thought it was in some kind of oscillating path so I don't know why it is not DNE.

Answer 8

(-1)^n / n
If n is odd, numerator is -1.
If n is even, numerator is +1.
So the numerator is a finite constant (either -1 or +1).
But the denominator is n which gets very large as n approaches infinity.
So the quotient approaches 0.

Answer 9

oh waittttt. oops. i get it now. thanks!

Answer 10

If n = 1000, (-1)^n / n = 1 / 1000 = 0.001
If n = 1001, (-1)^n / n = -1 / 1001 approx = -0.001

Keep increasing n and you will find it approaches zero.

Answer 11

You are welcome.