Chapter 8: Infinite Sequences and Series
Section 8.2: Series
Show that Leibniz' theorem (see Table 8.2.2) on the convergence of alternating series applies to the series ∑n=1∞−1n+1n.
Obtain the first few, but graph the first 50, partial sums.
If Sk is the partial sum of the first k terms, what value of k will guarantee that the error in Sk is no worse than 10−3?
In the alternating series ∑n=1∞−1n+1n, an=1/n defines a sequence of positive terms that decrease monotonically to zero. Hence the conditions of Leibniz' theorem hold, and the sequence converges conditionally.
Maple is unable to obtain the sum of the series
Write the alternating harmonic series.
Context Panel: Evaluate and Display Inline
∑n=1∞−1n+1n = ∑n=1∞−1n+1n
Maple is unable to obtain an expression for the kth partial sum
Control-drag the series and change ∞ to k.
Context Panel: Assign to a Name≻S[k]
Sk=∑n=1k−1n+1n→assign as functionS
Display the first few partial sums
Type Sk and press the Enter key.
Context Panel: Sequence≻k
In the resulting dialog box, set k=1 to k=15
Figure 8.3.17(a) shows the convergence of the first 50 members of the sequence of partial sums.
use plots in
Figure 8.2.17(a) Convergence of Sk to S
The question amounts to this: what is the smallest value of k for which 1/k+1 is less than 10−3? Hence, solve the equation 1/k+1=10−3 for k+1=106, so the first 999,999 terms have to be summed to be sure the result has an error no worse than 10−3. This series converges very slowly!
The desired accuracy might be achieved with fewer terms, but using the estimate of error in these notes, it would take nearly one million terms to guarantee that accuracy.
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