# How do you determine the convergence or divergence of Sigma ((-1)^(n+1))/(2n-1) from [1,oo)?

##### 2 Answers
Nov 20, 2016

Alternating series, which alternate between having positive and negative terms, often come in the forms ${\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n} {a}_{n}$ or ${\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n + 1} {a}_{n}$. The only difference between these two 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 ${\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n} {a}_{n}$ or ${\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n + 1} {a}_{n}$, the sum will converge if both:

• ${\lim}_{n \rightarrow \infty} {a}_{n} = 0 \text{ "" }$ (the sequence approaches $0$)
• ${a}_{n} \ge {a}_{n + 1} \text{ "" }$ (the sequence is decreasing)

So, for ${\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n + 1} / \left(2 n - 1\right)$, we see that the sequence being alternated is ${a}_{n} = \frac{1}{2 n - 1}$.

We see that ${\lim}_{n \rightarrow \infty} {a}_{n} = {\lim}_{n \rightarrow \infty} \frac{1}{2 n - 1} = {\lim}_{n \rightarrow \infty} \frac{\frac{1}{n}}{2 - \frac{1}{n}} = 0$. We also can see that as $n \rightarrow \infty$, the denominator of $\frac{1}{2 n - 1}$ will steadily grow, so $\frac{1}{2 n - 1}$ will steadily decrease.

We can also show that ${a}_{n} \ge {a}_{n + 1}$ by actually setting up that inequality:

$\frac{1}{2 n - 1} \ge \frac{1}{2 \left(n + 1\right) - 1} \implies \frac{1}{2 n - 1} \ge \frac{1}{2 n + 1}$

And then by cross-multiplying and solving this inequality, which is tedious, we can show that this is always true (at least for $n > 1$).

Anyway, we've showed that ${\lim}_{n \rightarrow \infty} {a}_{n} = 0$ and ${a}_{n} \ge {a}_{n + 1}$, so we can conclude that ${\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n + 1} / \left(2 n - 1\right)$ converges.

Nov 20, 2016

Arranging and adding successive terms

$\frac{1}{2 k - 1} - \frac{1}{2 \left(k + 1\right) - 1} = \frac{1}{2 k - 1} - \frac{1}{2 k + 1} = \frac{2}{{\left(2 k\right)}^{2} - 1}$

so

${\sum}_{k = 1}^{\infty} {\left(- 1\right)}^{k + 1} / \left(2 k - 1\right) = {\sum}_{k = 0}^{\infty} \frac{2}{{\left(2 \left(2 k + 1\right)\right)}^{2} - 1}$.

and we have

${\sum}_{k = 0}^{\infty} \frac{2}{{\left(2 \left(2 k + 1\right)\right)}^{2} - 1} < {\sum}_{k = 0}^{\infty} \frac{2}{2 \left(2 k + 1\right)} ^ 2 < 2 {\sum}_{k = 1}^{\infty} \frac{1}{k} ^ 2 = 2 {\pi}^{2} / 6$

so

${\sum}_{k = 1}^{\infty} {\left(- 1\right)}^{k + 1} / \left(2 k - 1\right)$ converges.