Exercise · ZW-NUMERICAL-SERIES-010 · ★★★ Advanced
Which statement about \(\sum a_n\) is true?
Choose one option.
Show answer
- B. If the series converges, then \(a_n\to0\); the converse is false
Explanation
Convergence implies \(a_n\to0\) since \(a_n=s_n-s_{n-1}\). The converse fails: \(\sum\frac1n\) diverges although \(\frac1n\to0\).
Review the concept: Numerical series
A series is the sequence of partial sums \(s_N\) of a sequence of terms \(a_n\); it converges if \(s_N\) has a finite limit, and that limit is its sum.
\[s_N=\sum_{n=1}^{N}a_n,\qquad \sum_{n=1}^{\infty}a_n=\lim_{N\to\infty}s_N\]
More practice problems: numerical series
- ZW-NUMERICAL-SERIES-001 Compute ∑_n=0^∞(tfrac12)^n.
- ZW-NUMERICAL-SERIES-002 Compute ∑_n=1^∞(1)/(n(n+1)).
- ZW-NUMERICAL-SERIES-003 For the series ∑_n=1^∞1n^2, compute the partial sum s_3.
- ZW-NUMERICAL-SERIES-004 What happens with ∑_n=0^∞(-1)^n?
- ZW-NUMERICAL-SERIES-005 Which of these series diverges because its general term does not tend to zero?
- ZW-NUMERICAL-SERIES-006 Compute ∑_n=1^∞(3)/(4^n).
- ZW-NUMERICAL-SERIES-007 Compute ∑_n=2^∞(1)/(n^2-1).
- ZW-NUMERICAL-SERIES-008 Compute ∑_n=0^∞(-tfrac23)^n.