Exercise · ZW-NUMERICAL-SERIES-002 · ★☆☆ Basic
Compute \(\displaystyle\sum_{n=1}^{\infty}\frac{1}{n(n+1)}\).
Choose one option.
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- C. \(1\)
Explanation
Since \(\frac1{n(n+1)}=\frac1n-\frac1{n+1}\), the partial sum is \(s_N=1-\frac1{N+1}\), which tends to \(1\).
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-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.
- ZW-NUMERICAL-SERIES-009 The partial sums of a series are s_N=(2N)/(N+1). What is the term a_3?