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Question: How do you convergence or divergence of \[\sum{\dfrac{{{\left( -1 \right)}^{n+1}}}{2n-1}}\] from \[\...

How do you convergence or divergence of ∑(−1)n+12n−1\sum{\dfrac{{{\left( -1 \right)}^{n+1}}}{2n-1}} from [1,∞)\left[ 1,\infty \right)?

Explanation

Solution

To find that given function ∑(−1)n+12n−1\sum{\dfrac{{{\left( -1 \right)}^{n+1}}}{2n-1}} is convergence or divergence. Let us consider the given function as equation (1). Then by using alternate series tests we can easily find the given function converges or diverges. If the given function satisfies the conditions lim⁡n→∞an=0\displaystyle \lim_{n \to \infty }{{a}_{n}}=0 and an≥an+1{{a}_{n}}\ge {{a}_{n+1}} then our function will converge.

Complete step-by-step answer:
For the given problem we are given to find convergence or divergence ∑(−1)n+12n−1\sum{\dfrac{{{\left( -1 \right)}^{n+1}}}{2n-1}} from [1,∞)\left[ 1,\infty \right).
Let us consider the given equation as equation (1).
L=∑(−1)n+12n−1............(1)L=\sum{\dfrac{{{\left( -1 \right)}^{n+1}}}{2n-1}}............\left( 1 \right)
As we know that alternating series, which alternate between having positive and negative terms, often come in the forms ∑n=1∞(−1)nan\sum\limits_{n=1}^{\infty }{{{\left( -1 \right)}^{n}}{{a}_{n}}} or∑n=1∞(−1)n+1an\sum\limits_{n=1}^{\infty }{{{\left( -1 \right)}^{n+1}}{{a}_{n}}}. The only difference between these terms is which terms are positive and which are negative.
Leibniz's rule, or the alternating series test, can be used to determine if one of these series converges or not.
For an alternating series such as ∑n=1∞(−1)nan\sum\limits_{n=1}^{\infty }{{{\left( -1 \right)}^{n}}{{a}_{n}}} or ∑n=1∞(−1)n+1an\sum\limits_{n=1}^{\infty }{{{\left( -1 \right)}^{n+1}}{{a}_{n}}}, the sum will be converge if both:
lim⁡n→∞an=0\displaystyle \lim_{n \to \infty }{{a}_{n}}=0 and an≥an+1{{a}_{n}}\ge {{a}_{n+1}} .
So, for ∑(−1)n+12n−1\sum{\dfrac{{{\left( -1 \right)}^{n+1}}}{2n-1}} , we see that the sequence being alternated is an=12n−1{{a}_{n}}=\dfrac{1}{2n-1}.
We see that lim⁡n→∞an=lim⁡n→∞12n−1=lim⁡n→∞1n2−1n=0\displaystyle \lim_{n \to \infty }{{a}_{n}}=\displaystyle \lim_{n \to \infty }\dfrac{1}{2n-1}=\displaystyle \lim_{n \to \infty }\dfrac{\dfrac{1}{n}}{2-\dfrac{1}{n}}=0 . We can also see that as n→∞n \to \infty , the denominator will grow, so 12n−1\dfrac{1}{2n-1} will steadily decrease.
We can also show that an≥an+1{{a}_{n}}\ge {{a}_{n+1}} by actually setting up that inequality:
12n−1≥12(n+1)−1⇒12n−1≥12n+1\dfrac{1}{2n-1}\ge \dfrac{1}{2\left( n+1 \right)-1}\Rightarrow \dfrac{1}{2n-1}\ge \dfrac{1}{2n+1}
And then by cross-multiplying and solving this inequality, which is tedious, we can show that this is always true.
Anyway, we've showed that lim⁡n→∞an=0\displaystyle \lim_{n \to \infty }{{a}_{n}}=0 and an≥an+1{{a}_{n}}\ge {{a}_{n+1}}, so we can conclude that ∑(−1)n+12n−1\sum{\dfrac{{{\left( -1 \right)}^{n+1}}}{2n-1}} converges.

Note: We should know the concept of limits perfectly for doing this problem. Students may stuck at the point i.e. 12n−1≥12(n+1)−1⇒12n−1≥12n+1\dfrac{1}{2n-1}\ge \dfrac{1}{2\left( n+1 \right)-1}\Rightarrow \dfrac{1}{2n-1}\ge \dfrac{1}{2n+1}. So, students should understand the problem carefully. Students should avoid calculation mistakes while doing this problem.