Inequality and Infinite Sum of A Function

Question

Part 1

Have been given the following inequality,

$$\left|\frac 1 {2+a}\right|<1 $$

I am a but confused about how to solve it.

My Approach

I have used the following method, $$\left|\frac 1 {2+a}\right|<1 $$ $$-1<\frac{1}{(2+a)}, \qquad \frac{1}{(2+a)}<1$$

$$-(2+a)>1, \qquad 1>(2+a)$$

$$(2+a)<-1, \qquad 1>(2+a)$$

$$a<-1-2, \qquad 1-2>a$$

$$a<-3, \qquad -1>a$$

But then I get the following result

$$a<-3, \qquad -1>a$$

Which I am not sure if it is correct or not. I am just unsure about the step that I have done which was when I timed the fraction on the otherside, I turned the inequality because its given in the question that $a$ can be any number so hence it can also be negative, right? So is it right what I have done? Or is it wrong.

Part 2 For the given values of a, find the sum of the following,

$$\sum_{n=1}^\infty \frac 1 {(2+a)^n}$$

This one I am a bit lost.

Answer 1

For part 2 you have to calculate the sum of a geometric series and it is $\frac 1 {1-b}$ where $b=2+a$ (or $2-a$, I've not clear which is the correct one). In part 2 the start was correct but then $2-a$ turned in $2-a$ and maybe you forgot some $-$.

Answer 2

From $ -1 < \dfrac 1 {2+a}$ you cannot deduce that $-1\cdot(2+a) < 1$ unless you know that $2+a$ is positive, nor can you deduce that $-1\cdot(2+a) > 1$ unless you know that $2+a$ is negative.

But you can do this: \begin{align} -1 & < \frac 1 {2+a} \\[10pt] 0 & < \frac 1 {2+a} +1 \\[10pt] 0 & < \frac 1 {2+a} + \frac{2+a}{2+a} \\[10pt] 0 & < \frac{3+a}{2+a} \end{align} Then look at three intervals: $(-\infty,-3),\ (-3,-2),\ (-2,+\infty)$.

Answer 3

your inequality is equivalent to $$1<|2+a|$$ and $$a \ne -2$$ thus we have for $-2<a$ $$1<2+a$$ thus we get $$a<-1$$ and if $$-2>a$$ we get $$1<-(2+a)$$ or $$3<-a$$ which gives $$a<-3$$