Determine whether the fallowing series is absolutely convergent, conditionally convergent or divergent.?

Question

sum of (-1)^n*(n/n+1)^n from n=1 to infinity,
Give your reason(s) for your decision.

Answer 1

since it is an alternating series, you must prove the three points:

1) (n/n+1)^n is positive for all n

2) (n/n+1)^n > ((n+1)/(n+2))^(n+1) for all n
...this also means each succeeding term is less

3) lim (n/n+1)^n approaches infinity

fulfills all conditions
absolutely convergent

Answer 2

Actually, the above poster is completely wrong. This series diverges. In order for it to be convergent, the limit of the terms in the series as n approaches infinity would have to be zero, but in fact the limit does not exist ([n→∞]lim (n/(n+1))^n = 1/e, so (-1)^n (n/(n+1))^n would oscillate between values close to 1/e and -1/e, and thus would not stay close to any particular number). Therefore, the series does not converge.

Proof that [n→∞]lim (n/(n+1))^n = 1/e:

Note that since the natural logarithm is a continuous function, ln [n→∞]lim (n/(n+1))^n = [n→∞]lim ln (n/(n+1))^n, which is

[n→∞]lim n ln (n/(n+1))
[n→∞]lim ln (n/(n+1))/(1/n)

This is now in 0/0 form, so we may employ L'hopital's rule:

[n→∞]lim ((n+1)/n) * (n+1 - n)/(n+1)² / (-1/n²)

Simplifying:

[n→∞]lim -n/(n+1)

-1

Since ln [n→∞]lim (n/(n+1))^n = -1, it follows that [n→∞]lim (n/(n+1))^n = e^(-1) = 1/e, which was to be demonstrated.