Determine if absolutely convergent, conditionally convergent or divergent?

Question

1. n=1 to infinity ( n! / n^n)

Please show steps and explain thoroughly.

He_Who_Shall_Not_Be_Named · Accepted Answer

Absolutely convergent. By the ratio test:

lim(n->infinity) | a(n+1) / a(n) | =

lim(x->infinity) | (x+1)! * x^x / ( x! * (x+1)^(x+1) ) |

= lim(x->inf) | (x+1) * x^x / ( (x+1) * (x+1)^x ) |

= lim(x->inf) | x^x / (x+1)^x |

= lim(x->inf) | (x/(x+1))^x |

= 1 / lim(x->inf) | 1 / (x/(x+1))^x |

= 1 / lim(x->inf) | 1^x / (x/(x+1))^x |

= 1 / lim(x->inf) | ((x+1)/x)^x |

= 1 / lim(x->inf) | (1 + 1/x)^x |

= 1 / |e|

= 1/e < 1

1 < e ... check

Therefore, by the ratio test, the series converges absolutely.

Unknown · Answer

n=1 to inf (n!)/(n^n) =

n=1 to inf (1*2*3*.....*n*.....)/(n*n*n*n.......*n*.....)

n = 1 to inf (1/n*2/n*3/n.....n/n.........)

n = 1 1/n
n-2 = 2/n*1/n
n = 3 = 3/n*2/n*1/n
...
= 1/1 + 2/4 + 6/27 + 24/256 + 120/3125`

so 1/n^2 = 1/1 + 1/4 + 1/9 + 1/16 + 1/25...converges to pi^2/6

note that the convergent function is term wise larger than n!/n^n all terms n>=5

So since the sum 1/5^2 + 1/6^2 ,,,> 120/5^5 +720/46456

so by the comparison test our series converges absolutely for n>=5 the terms for n = 1 to 4 are finite in both series,

1, 1/4, 1/9, 1/16 and 1, 1/2 .2/9, 3/8,.3/32  hence the series converges absolutely.

Trending News