Series question . . .?

Question

Test the infinite series for convergence or divergence:

E [ (2)^(1/n) - 1 ]

Answer 1

The mathemagician had a brilliant idea, but unfortunately he made an algebraic mistake:

He wrote:

"Thus, also, lim n->∞ (1 + 1/n^(4/3))^(n^(4/3)) = e.

Taking the 3/4 root gives lim n->∞ (1 + 1/n^(4/3))^n = e^(3/4)."

This is a mistake, because (1 + 1/n^(4/3))^(n^(4/3)) IS NOT [(1 + 1/n^(4/3))^(n)](4/3). So, taking the 3/4 root does not lead to his conclusion. Just an algebraic mistake that anyone can make.

Actually, the series diverges. To see this, observe that, according to Taylor expansion, for every a>1 and every real x we have

a^x = 1 + x ln(a) + x^2/2 (ln(a))^2 ...+ x^n/n! (ln(a))^n....

Thefore, for x>0 we have a^x > 1 + x ln(a) => a^(1/x) > 1 + ln(a)/x. In particular, it follows that, for every integer n >=1, we have
2^(1/n) > 1 + ln(2)/n => 2^(1/n) -1 > ln(2)/n

Sum (ln(2)/n diverges, because the harmonic series Sum 1/n diverges. By comparisson, we conclude Sum (2^(1/n) -1) diverges.

Answer 2

EDIT: The answerer after me is correct; I made a mistake in my solution. The series diverges. Please refer to his solution instead.

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This is a tough one (unless I'm missing something). The series converges.

Use direct comparison with 1/n^(4/3).

Recall that lim n->∞ (1 + 1/n)^n = e.

Thus, also, lim n->∞ (1 + 1/n^(4/3))^(n^(4/3)) = e.

Taking the 3/4 root gives lim n->∞ (1 + 1/n^(4/3))^n = e^(3/4).

Since e^(3/4) > 2, then in particular, for large values of n, we have

(1 + 1/n^(4/3))^n > 2

Take the nth root:

1 + 1/n^(4/3) > 2^(1/n)

Subtract 1:

1/n^(4/3) > 2^(1/n) - 1.

Since 0 < 2^(1/n) - 1 < 1/n^(4/3) for large values of n, and ∑1/n^(4/3) converges, then ∑(2^(1/n) - 1) converges as well.

Answer 3

just because lim n=>oo gives 0 does not mean converges

Ratio Test:
lim n=>oo I 2^(1/(n+1)) -1 divide by 2^(1/n) -1 I
Use L'Hopital's Rule
lim n=>oo I 2^(1/(n+1)) *(-1/(n+1)^2) divide by 2^(1/n) * (-1/n^2)
lim n=>oo I 2^ (1/n+1 - 1/n) * n^2 / (n+1)^2 I
evaluate limit 2^0 = 1
INCONCLUSIVE.. DARN!

Answer 4

This should DIVERGE, since after a while, the 2^(1/n) part of each term should approach zero, leaving n(-1) where n goes to infinity.

Answer 5

limit as it goes to infinity is 0, so it converges. i think. well u can just plug in values and see they arent increasing or alternating, so by common sense converges. as for the sum well good night.