# Thread: Complete Interval of Convergence Question(s)

1. ## Complete Interval of Convergence Question(s)

Well, I just need a little help in these... There are parts that I can't completely understand:

Find the complete interval of convergence for each series

1. $\displaystyle \sum_{n = 1}^{\infty} \frac{(-1)^{n + 1}(x + 5)^{2n}}{2n}$

For this one... All I could do was set up the limit for the ratio test:

$\displaystyle L = \lim_{n\to \infty} \left| \frac{(-1)^{n + 2}(x + 5)^{2n + 2}}{(2n + 2)} * \frac{2n}{(-1)^{n + 1}(x + 5)^{2n}}\right|$

2. $\displaystyle \sum_{n = 0}^{\infty} \frac{\ln{(n + 1)}x^n}{(n + 1)}$

For this one, I did a bit of work... But I need to know something:

$\displaystyle L = \lim_{n\to \infty} \left| \frac{ln(n + 2)x^{n + 1}}{(n + 2)} * \frac{(n + 1)}{ln(n + 1)x^n}\right|$

$\displaystyle = \lim_{n\to \infty} \left| \frac{ln(n+2)(n+1)x}{ln(n+1)(n + 2)}\right|$

$\displaystyle = |x|\lim_{n\to \infty} \left| \frac{(ln(n + 2))^{n + 1}}{(ln(n + 1))^{n + 2}}\right|$

This supposedly equals:

$\displaystyle L = |x|$

But I don't really know how the limit gets to 1.

Any help is appreciated.

2. Originally Posted by Aryth Well, I just need a little help in these... There are parts that I can't completely understand:

Find the complete interval of convergence for each series

1. $\displaystyle \sum_{n = 1}^{\infty} \frac{(-1)^{n + 1}(x + 5)^{2n}}{2n}$

For this one... All I could do was set up the limit for the ratio test:

$\displaystyle L = \lim_{n\to \infty} \left| \frac{(-1)^{n + 2}(x + 5)^{2n + 2}}{(2n + 2)} * \frac{2n}{(-1)^{n + 1}(x + 5)^{2n}}\right|$

2. $\displaystyle \sum_{n = 0}^{\infty} \frac{\ln{(n + 1)}x^n}{(n + 1)}$

For this one, I did a bit of work... But I need to know something:

$\displaystyle L = \lim_{n\to \infty} \left| \frac{ln(n + 2)x^{n + 1}}{(n + 2)} * \frac{(n + 1)}{ln(n + 1)x^n}\right|$..
$\displaystyle = \lim_{n\to \infty} \left| \frac{ln(n+2)(n+1)x}{ln(n+1)(n + 2)}\right|$

$\displaystyle = |x|\lim_{n\to \infty} \left| \frac{(ln(n + 2))^{n + 1}}{(ln(n + 1))^{n + 2}}\right|$

This supposedly equals:

$\displaystyle L = |x|$

But I don't really know how the limit gets to 1.

Any help is appreciated.
For the first one you should have $\displaystyle \bigg|\frac{(x+5)^{2n+1}}{2n+1}\cdot\frac{2n}{(x+5 )^{2n}}\bigg|<1$...therefore $\displaystyle \lim_{n \to {\infty}}\bigg|\frac{2n(x+5)}{2n+1}\bigg|<1$ so $\displaystyle |x+5|<1\Rightarrow{-6\leq{x}\leq{-4}}$..I got the inclusion by testing those points for convergence in the original series

3. Thanks that helped quite a bit.

I'm going to go ahead and finish that one and see if the rest of it is right:

$\displaystyle \text{Int. of Conv} = (-6, -4)$

For x = -6:

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

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

$\displaystyle \text{First few terms:}\frac{1}{2} - \frac{1}{4} + \frac{1}{6} - ...$

The alternating series converges because:

1. $\displaystyle |a_{n + 1}| < |a_n|$

2. $\displaystyle \lim_{n\to \infty} a_n = 0$

Therefore $\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n + 1}(x + 5)^{2n}}{2n}$ converges at x = -6.

For x = -4:

This would lead to the same series, meaning that the sum converges for x = -4 as well.

$\displaystyle \text{Complete Interval of Convergence} = [-6, -4]$

Is that it?

4. Originally Posted by Aryth Thanks that helped quite a bit.

I'm going to go ahead and finish that one and see if the rest of it is right:

$\displaystyle \text{Int. of Conv} = (-6, -4)$

For x = -6:

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

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

$\displaystyle \text{First few terms:}\frac{1}{2} - \frac{1}{4} + \frac{1}{6} - ...$

The alternating series converges because:

1. $\displaystyle |a_{n + 1}| < |a_n|$

2. $\displaystyle \lim_{n\to \infty} a_n = 0$

Therefore $\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n + 1}(x + 5)^{2n}}{2n}$ converges at x = -6.

For x = -4:

This would lead to the same series, meaning that the sum converges for x = -4 as well.

$\displaystyle \text{Complete Interval of Convergence} = [-6, -4]$

Is that it? great job!

5. Originally Posted by Aryth Well, I just need a little help in these... There are parts that I can't completely understand:

Find the complete interval of convergence for each series

1. $\displaystyle \sum_{n = 1}^{\infty} \frac{(-1)^{n + 1}(x + 5)^{2n}}{2n}$

For this one... All I could do was set up the limit for the ratio test:

$\displaystyle L = \lim_{n\to \infty} \left| \frac{(-1)^{n + 2}(x + 5)^{2n + 2}}{(2n + 2)} * \frac{2n}{(-1)^{n + 1}(x + 5)^{2n}}\right|$

2. $\displaystyle \sum_{n = 0}^{\infty} \frac{\ln{(n + 1)}x^n}{(n + 1)}$

For this one, I did a bit of work... But I need to know something:

$\displaystyle L = \lim_{n\to \infty} \left| \frac{ln(n + 2)x^{n + 1}}{(n + 2)} * \frac{(n + 1)}{ln(n + 1)x^n}\right|$

$\displaystyle = \lim_{n\to \infty} \left| \frac{ln(n+2)(n+1)x}{ln(n+1)(n + 2)}\right|$

$\displaystyle = |x|\lim_{n\to \infty} \left| \frac{(ln(n + 2))^{n + 1}}{(ln(n + 1))^{n + 2}}\right|$

This supposedly equals:

$\displaystyle L = |x|$

But I don't really know how the limit gets to 1.

Any help is appreciated.
For the second one you have $\displaystyle \lim_{n \to {\infty}}\bigg|\frac{ln(n+2)x^{n+1}}{n+2}\cdot\fra c{n+1}{x^{n}ln(n+1)}\bigg|<1$...so $\displaystyle |x|<1$ the way I solved this was by looking at the limit as n gets infinitely large the +1 won't matter...so I extracted teh x and called the limit one...and test for convergence at x=-1,x=1 and I will crique you

6. These series are quite confusing, our teacher never told us the behavior of series with natural logarithms, for example:

When x = -1 it gives an alternating series, supposedly it converges, but I can't prove it necessarily since the first term of it is 0 and the second term is $\displaystyle -\frac{ln(2)}{2}$. Of course only the absolute value of it matters, but the terms in the sequence decrease only when $\displaystyle n \geq 3$... Does that still imply convergence?

7. Originally Posted by Aryth These series are quite confusing, our teacher never told us the behavior of series with natural logarithms, for example:

When x = -1 it gives an alternating series, supposedly it converges, but I can't prove it necessarily since the first term of it is 0 and the second term is $\displaystyle -\frac{ln(2)}{2}$. Of course only the absolute value of it matters, but the terms in the sequence decrease only when $\displaystyle n \geq 3$... Does that still imply convergence?
Yes it does...lets say the series for $\displaystyle n\geq{3}$ sums to some value denoted b(we know it sums to soem arbitrary value since as you proved the series converges for $\displaystyle n\geq{3}$)? then the real sum of the entire series is $\displaystyle a_0+a_2+a_3+b$ which is obviously convergent

8. Oh yeah... >_>

Haha.

Anyway, when x = 1. All I did was compare it to the harmonic series, which is less than the series we have, and since the harmonic series diverges, then our series diverges as well.

Giving us:

$\displaystyle \text{Complete Interval of Convergence} = [-1, 1)$

That right?

9. Originally Posted by Aryth Oh yeah... >_>

Haha.

Anyway, when x = 1. All I did was compare it to the harmonic series, which is less than the series we have, and since the harmonic series diverges, then our series diverges as well.

Giving us:

$\displaystyle \text{Complete Interval of Convergence} = [-1, 1)$

That right?
an easier way is this $\displaystyle \sum_{n=0}^{\infty}\frac{ln(n+1)}{n+1}$...convergence can be tested this way $\displaystyle \int_1^{\infty}\frac{ln(x+1)}{x+1}dx=[ln(x+1)]^2\bigg|_1^{\infty}=\infty$ therefore it is divergent...and we can use this test becuase $\displaystyle a_n$ is monotonically decreasing...

10. Originally Posted by Aryth Oh yeah... >_>

Haha.

Anyway, when x = 1. All I did was compare it to the harmonic series, which is less than the series we have, and since the harmonic series diverges, then our series diverges as well.

Giving us:

$\displaystyle \text{Complete Interval of Convergence} = [-1, 1)$

That right?
Yes.

11. Thanks guys. The help was much needed and much appreciated.

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