Solving Equation for Real Solutions

[springo]

Homework Statement

I have to find the number of (real) solutions to the following equation and place them in intervals of size 1.
$$\sum_{n=1}^{\infty}\frac{x^n+n^2+n}{x^{n+1}\cdot n\cdot(n+1)}=1$$

Homework Equations

The Attempt at a Solution

I have been able to prove that x > 1, because the series does not converge if x < 1. I was thinking of finding how much the series is, but I don't think I can do that (or at least I don't know how to).

 

[Mark44]

This isn't an equation, so it doesn't make sense to look for solutions.

Are you asking for what values of x (as intervals) this series converges?

 

[springo]

Yes, that's what I'm asking. Thanks for any help.

 

[Mark44]

Can you show what you did to establish that the series diverges for x < 1 and that it converges for x > 1?

 

[springo]

If x < 1:
$$\lim_{n\rightarrow\infty}\frac{x^n+n^2+n}{x^{n+1}\cdot n\cdot(n+1)}=\lim_{n\rightarrow\infty}\frac{n\cdot(n+1)}{x^{n+1}\cdot n\cdot(n+1)}=\lim_{n\rightarrow\infty}\frac{1}{x^{n+1}}=\infty$$
So the series does not converge.

However if x > 1:
$$\lim_{n\rightarrow\infty}\frac{x^n+n^2+n}{x^{n+1}\cdot n\cdot(n+1)}=\lim_{n\rightarrow\infty}\frac{x^n}{x^{n+1}\cdot n\cdot(n+1)}=\lim_{n\rightarrow\infty}\frac{1}{x\cdot n\cdot(n+1)}=0$$
So the series might converge if x > 1.

$$\lim_{n\rightarrow\infty}n\cdot\left(1-\frac{a_{n+1}}{a_{n}}\right)=\lim_{n\rightarrow\infty}n\cdot\left(1-\frac{\frac{1}{x\cdot (n+1)\cdot(n+2)}}{\frac{1}{x\cdot n\cdot(n+1)}}\right)=\lim_{n\rightarrow\infty}n\cdot\left(1-\frac{n}{n+2}\right)=\lim_{n\rightarrow\infty}\frac{2n}{n+2}=2$$
Therefore the series converges if x > 1.

 

[springo]

Oh I just noticed for some reason the tex didn't display the equation properly...
I want to find the solutions so that the series equals to 1, but it didn't display it.

Sorry about that.

 

[Mark44]

If x < 1:
$$\lim_{n\rightarrow\infty}\frac{x^n+n^2+n}{x^{n+1}\cdot n\cdot(n+1)}=\lim_{n\rightarrow\infty}\frac{n\cdot(n+1)}{x^{n+1}\cdot n\cdot(n+1)}=\lim_{n\rightarrow\infty}\frac{1}{x^{n+1}}=\infty$$
So the series does not converge.

What's your justification for the expression after the first = sign? Why does x disappear from the numerator, but not the denominator? If you're cancelling, you can't do that, since the x term isn't a factor in the numerator.

However if x > 1:
$$\lim_{n\rightarrow\infty}\frac{x^n+n^2+n}{x^{n+1}\cdot n\cdot(n+1)}=\lim_{n\rightarrow\infty}\frac{x^n}{x^{n+1}\cdot n\cdot(n+1)}=\lim_{n\rightarrow\infty}\frac{1}{x\cdot n\cdot(n+1)}=0$$
So the series might converge if x > 1.

What's your justification for the expression after the first = sign? Why did the terms in n disappear from the numerator?

$$\lim_{n\rightarrow\infty}n\cdot\left(1-\frac{a_{n+1}}{a_{n}}\right)=\lim_{n\rightarrow\infty}n\cdot\left(1-\frac{\frac{1}{x\cdot (n+1)\cdot(n+2)}}{\frac{1}{x\cdot n\cdot(n+1)}}\right)=\lim_{n\rightarrow\infty}n\cdot\left(1-\frac{n}{n+2}\right)=\lim_{n\rightarrow\infty}\frac{2n}{n+2}=2$$
Therefore the series converges if x > 1.

 

[Dick]

Sure the series converges for x>1. Break it up into 1/(x*n*(n+1))+1/x^(n+1). Compute the sum of each series. One is geometric and one telescopes.

 

[springo]

If x < 1, xⁿ is 0 as n approaches infinity whereas n(n+1) approaches infinity, so we can take xⁿ out.
If x > 1, xⁿ approaches infinity much faster than n(n+1), so the following is true: xⁿ + n(n+1) ~ xⁿ.

OK, so I did split into geometric and telescope, added them back and found 1/(x-1) = 1. So x = 2, is that correct?

 

[Dick]

If x < 1, xⁿ is 0 as n approaches infinity whereas n(n+1) approaches infinity, so we can take xⁿ out.
If x > 1, xⁿ approaches infinity much faster than n(n+1), so the following is true: xⁿ + n(n+1) ~ xⁿ.

OK, so I did split into geometric and telescope, added them back and found 1/(x-1) = 1. So x = 2, is that correct?

That's what I got.

 

[springo]

Thanks a lot for your help!