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

##### 2 Answers
Dec 14, 2016

As:

$\frac{n + 1}{{\left(n + 1\right)}^{2} + 1} \le \frac{n}{{n}^{2} + 1}$

and:

${\lim}_{n \to \infty} \frac{n}{{n}^{2} + 1} = 0$

the series is convergent.

#### Explanation:

An alternative series:

${\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n} {a}_{n}$

converges if the succession $\left\{{a}_{n}\right\}$ is decreasing and converges to zero.

So we make the test:

${a}_{n + 1} \le {a}_{n}$

$\frac{n + 1}{{\left(n + 1\right)}^{2} + 1} \le \frac{n}{{n}^{2} + 1}$

$\frac{n + 1}{{\left(n + 1\right)}^{2} + 1} - \frac{n}{{n}^{2} + 1} \le 0$

$\frac{\left(n + 1\right) \left({n}^{2} + 1\right) - n \left({\left(n + 1\right)}^{2} + 1\right)}{\left({\left(n + 1\right)}^{2} + 1\right) \left({n}^{2} + 1\right)} \le 0$

As the denominator is always positive we can focus on the numerator:

( (n+1)(n^2+1)-n( (n+1)^2+1)) = n^3+n+n^2+1 -n (n^2+2n+1+1) = n^3+n+n^2+1 -n^3-2n^2-2n = -2n^2-2n +1 <=0

which is always true, for $n > 1$ so the first condition is satisfied.

The we check that:

${\lim}_{n \to \infty} {a}_{n} = {\lim}_{n \to \infty} \frac{n}{{n}^{2} + 1} = 0$

So both conditions are satisified and the series is convergent.

Dec 14, 2016

Another way that showing ${a}_{n}$ is decreasing on $n \in \left[1 , \infty\right)$ other than showing that ${a}_{n + 1} < {a}_{n}$ is to find the derivative of ${a}_{n}$.

$\frac{d}{\mathrm{dx}} \left(\frac{x}{{x}^{2} + 1}\right) = \frac{\left(\frac{d}{\mathrm{dx}} x\right) \left({x}^{2} + 1\right) - x \left(\frac{d}{\mathrm{dx}} \left({x}^{2} + 1\right)\right)}{{x}^{2} + 1} ^ 2$

$\textcolor{w h i t e}{\frac{d}{\mathrm{dx}} \left(\frac{x}{{x}^{2} + 1}\right)} = \frac{1 \left({x}^{2} + 1\right) - x \left(2 x\right)}{{x}^{2} + 1} ^ 2$

$\textcolor{w h i t e}{\frac{d}{\mathrm{dx}} \left(\frac{x}{{x}^{2} + 1}\right)} = \frac{1 - {x}^{2}}{{x}^{2} + 1} ^ 2$

Examining the sign of the derivative, we see that the denominator is always positive. Thus the sign of the derivative as a whole is dependent on the sign of the numerator. When $x > 1$, the numerator is negative, so the derivative is negative.

A negative derivative shows a decreasing function. Thus, ${a}_{n}$ is decreasing on $n \in \left[1 , \infty\right)$.

Using this in conjunction with the fact that

${\lim}_{n \rightarrow \infty} {a}_{n} = {\lim}_{n \rightarrow \infty} \frac{n}{{n}^{2} + 1} = {\lim}_{n \rightarrow \infty} \frac{\frac{1}{n}}{1 + \frac{1}{n} ^ 2} = \frac{0}{1 + 0} = 0$,

we can claim that ${\sum}_{n = 1}^{\infty} \frac{{\left(- 1\right)}^{n} n}{{n}^{2} + 1}$ is convergent through the alternating series test.

We can go on to note that ${\sum}_{n = 1}^{\infty} \frac{n}{{n}^{2} + 1}$ is divergent through limit comparison with the divergent series ${\sum}_{n = 1}^{\infty} \frac{1}{n}$, so we can claim that ${\sum}_{n = 1}^{\infty} \frac{{\left(- 1\right)}^{n} n}{{n}^{2} + 1}$ is conditionally convergent.