State the alternating series test. Deduce that the series P = \(\Sigma_{n=1}^∞ (-1)^n {1\over\sqrt{n}}\) converges. Is this series absolutely convergent? Justify your answer. Find a divergent series which has the same terms (−1)n \({1\over\sqrt{n}}\) taken in a different order. You should justify the divergence. [You may use the comparison test, provided that you accurately state it.]
lim n-> ∞ b n = 0 lim n-> ∞ \({1\over\sqrt{n}}\) = 0 {bn} is a decreasing sequence. \({1\over\sqrt{n}}\)- \({1\over\sqrt{n+1}}\)>0
As the two conditions are met therefore we can say that the series P = \(\Sigma_{n=1}^∞ (-1)^n {1\over\sqrt{n}}\) is convergent.
In mathematics, an infinite series of numbers is said to converge absolutely (or to be absolutely convergent) if the sum of the absolute values of the summands is finite.
an < an +|an | < 2|an |
If \(\Sigma_{n=1}^∞ |a_n|\) is convergent then \(\Sigma_{n=1}^∞ 2|a_n|\) is also convergent.
Using the comparison test then \(\Sigma_{n=1}^∞ a_n + |a_n|\) is also convergent
\(\Sigma_{n=1}^∞ a_n\) = \(\Sigma_{n=1}^∞ a_n + |a_n|\) - \(\Sigma_{n=1}^∞ |a_n|\)
This series is the difference of two convergent series so is also a convergent series.
It can be shown using Taylor's series
ln(1+x)= 1 - x/2 + x/3 - x/4 +
Hence ln(2) = 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + 1/7 - 1/8 + .
ln(2) = (1 - 1/2) - (1/4) + (1/3 - 1/6) + 1/8 + (-1/5+1/10) -(1/12) + (1/7 - 1/14)
ln(2) = (1/2) - (1/4) + (-1/6) + 1/8 + (-1/10) + 1/14
ln(2) = 1/2ln(2)
We have shown that the series 1 - 1/2 + 1/3 - 1/4 converges both to ln(2) and 1/2ln(2)
ln(2) = 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + 1/7 - 1/8 + .
1/2ln(2) = 1/2 - 1/4 + 1/6 - 1/8 + 1/10
Adding the series for ln(2) and 1/2ln(2) leads to:
3/2ln(2) = 1 + [-1/2+1/2]+ [1/3] +[-1/4- 1/4]+1/5+[-1/6+1/6]+1/7+[-1/8 - 1/8] + ..
This simplifies to
3/2ln(2) = 1 + [1/3] +[-1/2]+1/5+[1/8]+1/7+[-1/4] + ..
Now we have shown that the series 1 - 1/2 + 1/3 - 1/4 also converges to 3/2ln(2). In fact, this series can be shown to converge to whatever you want it to be.
This series is an alternating series which satisfies the conditions to converge but is not absolutely convergent.
The p-value is -1 which is not greater than 1 which is required for convergence.
In contrast, a series that is absolutely convergent will converge to one finite value (Arfken,2013).
If the infinite series \(\Sigma b_n\) converges and 0< a n< b n for all sufficiently large n then \(\Sigma a_n\) also converges.
If the infinite series \(\Sigma a_n\) diverges and 0< a n < b n for all sufficiently large n then \(\Sigma b_n\) also diverges.
\(-{1\over\sqrt{1}} +{1\over\sqrt{2}}-{1\over\sqrt{3}}+{1\over\sqrt{4}}+....\)
\([-{1\over\sqrt{1}}+{1\over\sqrt{2}}] + [-{1\over\sqrt{3}}-{1\over\sqrt{5}}+{1\over\sqrt{4}}]+[-{1\over\sqrt{7}}-{1\over\sqrt{9}}-{1\over\sqrt{11}}+{1\over\sqrt{6}}]....\)
>-0.5 + -0.5 + -0.5 +.. = -∞
Using the comparison test, we have shown that the alternating series also diverges.