Testing for convergence
Sum of n!/(n^n) starting at n=1 & to infinity.thanks for your help
Sum of n!/(n^n) starting at n=1 & to infinity.thanks for your help
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Use the ratio test for these questions.
1) r = lim(n→∞) [(n+1)^2 / (n+1)!] / [n^2 / n!]
......= lim(n→∞) (n+1)^2 / [n^2 * (n+1)]
......= lim(n→∞) (n+1)/n^2
......= 0.
Since r = 0 < 1, this series converges.
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2) r = lim(n→∞) [(n+1)! / (n+1)^(n+1)] / [n! / n^n]
......= lim(n→∞) (n+1) * n^n / (n+1)^(n+1)
......= lim(n→∞) (n/(n+1))^n
......= lim(n→∞) 1/((n+1)/n)^n
......= lim(n→∞) 1/(1 + 1/n)^n
......= 1/e, via limit definition of e.
Since r = 1/e < 1, this series converges.
I hope this helps!
1) r = lim(n→∞) [(n+1)^2 / (n+1)!] / [n^2 / n!]
......= lim(n→∞) (n+1)^2 / [n^2 * (n+1)]
......= lim(n→∞) (n+1)/n^2
......= 0.
Since r = 0 < 1, this series converges.
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2) r = lim(n→∞) [(n+1)! / (n+1)^(n+1)] / [n! / n^n]
......= lim(n→∞) (n+1) * n^n / (n+1)^(n+1)
......= lim(n→∞) (n/(n+1))^n
......= lim(n→∞) 1/((n+1)/n)^n
......= lim(n→∞) 1/(1 + 1/n)^n
......= 1/e, via limit definition of e.
Since r = 1/e < 1, this series converges.
I hope this helps!