If the sequence is convergent, find the value of the limit.?

Question

A_n = n^3/n!

I think I need to use the squeeze th...

Answer 1

The first two answers are wrong: the first because according to his reasoning, since inputting any natural number n in n/(n+1) yields a greater number in the denominator than in the numerator then its limit is zero, whereas it actually is 1, and the second one because while applying the ratio's test he missed to see the limit is zero and not 1/(n+1)...
In fact the sequence converges to zero, and hereby two proofs:

First proof: Using the squeeze theorem for n > 5:

0 <= n^3/n! <= n^3/[(n-3)(n-2)(n-1)n] <= n^3/(n-3)^4 --> 0

Second proof: look at the series with general term n^3/n!, and appply the ratio's test to it:

[(n+1)^3/(n+1)!]*[n!/n^3] = (n +1)^2/n^3 --> 0 ==> the infinite series converge, and thus the general term's sequence MUST converge to zero.

Regards
Tonio

Answer 2

Like Dr D showed using the ratio test, this sequence converges to 0. And, since lim A_(n+1)/A(n) = 0, it also follows the series Sum (n= 1 ,oo) n^3/n! converges. Here, he made a mistake.

Answer 3

You could use the ratio test.
n+1 th term = (n+1)^3 / (n+1)!
ratio = (n+1)^3 / n^3 * n! / (n+1)!
which tends to 1 / (n+1) as n tends to infinity.

The sequence is convergent, however the sum of the series is not. The sequence converges to zero.

**EDIT**
Bertrando, check out the last line before the edit. It's in english.

Answer 4

This series is the series (one million, 0, -one million, 0, one million, 0, -one million, 0.....). It has a subsequence that converges to one million, yet another that converges to 0 and a third one that converges to -one million. subsequently, the series would not suggestions-set something, it diverges, nevertheless it relatively is bounded.

Answer 5

whatever values you put on n the denominator n! will give the higher result, concluding this limit simply approaches 0.