# Series convergence question

1. Nov 7, 2005

### Bob19

I have the following series:
$$S = \sum _{n=0} ^{\infty} 4^n (x+2)^n$$
Is that the same as $$4^n \sum_{n=0} ^{\infty} (x+2)^n = 4^n ((x+2) + (x+3) + \cdots + (x+n))$$ ???
Best Regards Bob

Last edited: Nov 7, 2005
2. Nov 7, 2005

### Tzar

No... it looks clearly different. What makes you think that?

3. Nov 7, 2005

### Bob19

Okay thank You I just had to be sure,

Taking the first sum into account:

$$S = \sum _{n=0} ^{\infty} 4^n (x+2)^n$$

which method do I use to show for which values of x the series converges?

Best Regards,

Bob

4. Nov 7, 2005

### HallsofIvy

Staff Emeritus
Either the ratio test or the root test will work. Typically, the ratio test is easier:

$$\sum_{n=0}^\infty}a_n$$, an positive, converges if
$$lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}$$ exists and is less than 1.
Here,
$$\left|a_n\right|= \left|4^n(x+2)^n\right|= \left|(4(x+2))^n\right|$$
$$\left|a_{n+1}\right|= \left|(4(x+2))^{n+1}\right|$$
so
$$\left|\frac{a_{n+1}}{a_n}\right|= \left|4(x+2)\right|$$

That will be less than 1 provided |x+2|< 1/4. In other words, for -2-1/4< x< -2+ 1/4 or -9/4< x< -7/4. Of course, you will need to check the endpoints.

For this example, since we have that "n" power, the ratio test is even easier.
$$\sum_{n=0}^\infty}a_n$$, an positive, converges if
$$lim_{n\rightarrow\infty}^n\sqrt{a_n}$$ exists and is less than 1.
$$^n\sqrt{\left|(4(x+2)^n\right|}= \left|4(x+2)\right|$$
so we must, again, have 4|x+2|< 1.

Actually, it would be much easier to just note that this is a geometric series with common ratio 4(x+ 2)!

Last edited: Nov 7, 2005
5. Nov 7, 2005

### Bob19

Then the values for which S converges are these:
-2-1/4< x< -2+ 1/4 or -9/4< x< -7/4 ?

Or is it the end points?

Best Regards,
Bob

6. Nov 7, 2005

### HallsofIvy

Staff Emeritus
What I said was "a series converges if the sequence of ratios $\frac{a_{n+1}}{a_n}[/tex] converges to a number less than 1 or if the sequence of roots [itex]^n\sqrt{a_n}$ converges to a number less than 1." What I should have said, but didn't, was that the series diverge if those sequence either diverge or converge to a number larger than 1. If the sequences converge to 1- either can happen. Typically one end point gives an alternating series that converges (by the alternating series test) and the other gives a series that does not converge- although it is possible for the series to converge at both end points. That was shy I said "Of course, you will need to check the endpoints."

7. Nov 8, 2005

### Bob19

Okay thank you,

the values for which the series S converges is that then that $$x \leq 2$$ cause these are the only values which makes the inequality true?

Best Regards,

Bob