{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

11.4.misc1-eg

# 11.4.misc1-eg - n n 3(or 1 n 2 and that the series is...

This preview shows page 1. Sign up to view the full content.

Example Use the Direct Comparison Test to determine if the series converges or diverges. X n =1 ln n ( n 3 + 1) . To use the Direct Comparison test to show convergence, we need to find a convergent series X n =1 b n for which ln n ( n 3 + 1) b n for all n sufficiently large. To use the Direct Comparison test to show divergence, we need to find a divergent series X n =1 c n for which ln n ( n 3 + 1) c n for all n sufficiently large. Notice that the numerator grows more slowly than n and the denominator grows with n 3 , so we suspect that the terms of the series will “behave like”
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: n n 3 (or 1 n 2 ) and that the series is likely convergent. To ﬁnd an upper bound b n , we bound the numerator above and the denominator below as follows: For all n ≥ 1, ln n < n and n 3 + 1 > n 3 from which it follows that 1 n 3 + 1 < 1 n 3 . Thus ln n n 3 + 1 < n n 3 = 1 n 2 . The series ∞ X n =1 1 n 2 is a convergent p –series, therefore by the Direct Comparison Test, both series converge....
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online