Mark M. answered 05/05/24
Retired math prof. Calc 1, 2 and AP Calculus tutoring experience.
an = n! / 94n
an+1 / an = [(n+1)! / 94n+1] [94n / n!] = (1/94)(n+1)
limn→∞ l an+1 / an l = (1/94)limn→∞ (n+1) = ∞
By the Ratio Test, the series diverges.
Mark M. answered 05/05/24
Retired math prof. Calc 1, 2 and AP Calculus tutoring experience.
an = n! / 94n
an+1 / an = [(n+1)! / 94n+1] [94n / n!] = (1/94)(n+1)
limn→∞ l an+1 / an l = (1/94)limn→∞ (n+1) = ∞
By the Ratio Test, the series diverges.
(n+1)!*94n/(n!*94n+1)
You should be able to finish it from here to conclude that the series diverges.
You should really be able to see this without the ratio test:
There are n terms in the numerator and n in the denominator.
For n≥94, each additional factor is greater than 1.
Therefore, the series cannot converge.
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