# Series (Convergence, Divergence)

• Mar 15th 2009, 09:09 PM
coolguy99
Series (Convergence, Divergence)
Was doing my homework, came up with 5 I could not get. :(

I'm not too knowledgeable about the special characters you guys use to make the math problems, so I'll do my best to use normal characters.. (keep in mind that each of these problems starts with E (sum), and unless otherwise noted, it starts at n=0 and goes to infinity)

#1: (x+3)^n / (2)^n, converge or diverge, and find the sum if it converges

I was thinking of using the Geometric test, but I don't know how that works when there is an x involved (1/3 converges, but does ((1+x)/3) converge?)

#2: 1/5 + 1/8 + 1/11 + 1/14 + 1/17

I wrote up the sum (1/(5 + 3n)) where n=0, goes to infinity

I imagine then that using the p-test, it diverges? or is this not valid?

#3 asks what value of p would make the series converge?

series is (n=1, to infinity): ln(p) / n^p

I'm pretty sure you need the integral test, but how do you take the integral of ln(x)/x^p?

#4: 1+ sin(n) / 10^n

Comparison test I'm guessing? But what do you compare it to? And If you do the inequality bit (0<1+sin(n)<2), how do you determine if 2/10^n converges?

#5 (n^2 - 5n) / (n^3 + n + 1)

Can you just use L'hopital? (2n - 5) / (2n^3 +1), which becomes 2 / 6n, which diverges?

As you can tell, I am not too good with these..

I tried to make it as clear as possible, but if you're confused about the way I conveyed a question, I'll gladly try to clarify further
• Mar 16th 2009, 06:49 AM
Jester
Quote:

Originally Posted by coolguy99
Was doing my homework, came up with 5 I could not get. :(

I'm not too knowledgeable about the special characters you guys use to make the math problems, so I'll do my best to use normal characters.. (keep in mind that each of these problems starts with E (sum), and unless otherwise noted, it starts at n=0 and goes to infinity)

#1: (x+3)^n / (2)^n, converge or diverge, and find the sum if it converges

I was thinking of using the Geometric test, but I don't know how that works when there is an x involved (1/3 converges, but does ((1+x)/3) converge?)

#2: 1/5 + 1/8 + 1/11 + 1/14 + 1/17

I wrote up the sum (1/(5 + 3n)) where n=0, goes to infinity

I imagine then that using the p-test, it diverges? or is this not valid?

#3 asks what value of p would make the series converge?

series is (n=1, to infinity): ln(p) / n^p

I'm pretty sure you need the integral test, but how do you take the integral of ln(x)/x^p?

#4: 1+ sin(n) / 10^n

Comparison test I'm guessing? But what do you compare it to? And If you do the inequality bit (0<1+sin(n)<2), how do you determine if 2/10^n converges?

#5 (n^2 - 5n) / (n^3 + n + 1)

Can you just use L'hopital? (2n - 5) / (2n^3 +1), which becomes 2 / 6n, which diverges?

As you can tell, I am not too good with these..

I tried to make it as clear as possible, but if you're confused about the way I conveyed a question, I'll gladly try to clarify further

#1: $\sum_{n=1}^\infty \left( \frac{x+3}{2} \right)^n$ converges if $\left| \frac{x+3}{2} \right| < 1$. If so, the sum is $S = \frac{a}{1-r}$ where a is the first term and $r =\frac{x+3}{2}$

#2 $\sum_{n=1}^\infty \frac{1}{3n+5}$ LCT with $\sum_{n=0}^\infty \frac{1}{n}$

#3 $\int \frac{\ln x}{x^p}\,dx$ use parts

#4: $\sum_{n=1}^\infty \frac{1+\sin n}{10^n} \le \sum_{n=1}^\infty \frac{2}{10^n}$ and use the direct comparison test.

#5 $\sum_{n=1}^\infty \frac{n^2-5n}{n^3+n+1}$ LCT with $\sum_{n=1}^\infty \frac{1}{n}$
• Mar 16th 2009, 07:35 AM
coolguy99
Quote:

Originally Posted by danny arrigo
#1: $\sum_{n=1}^\infty \left( \frac{x+3}{2} \right)^n$ converges if $\left| \frac{x+3}{2} \right| < 1$. If so, the sum is $S = \frac{a}{1-r}$ where a is the first term and $r =\frac{x+3}{2}$

I'm getting... (x+3) / (5-x) as a sum right now... assuming that a = (x+3)/(2) and r = (x+3)/(2^n-1)

Quote:

Originally Posted by danny arrigo
#2 $\sum_{n=1}^\infty \frac{1}{3n+5}$ LCT with $\sum_{n=0}^\infty \frac{1}{n}$

I got diverges, which I think is right, thanks :)

Quote:

Originally Posted by danny arrigo
#3 $\int \frac{\ln x}{x^p}\,dx$ use parts

I'm still having trouble integrating..
u = lnx
du = 1/x dx
dv = 1/(x^p)
v = (p+1)/(x^p+1)

?
• Mar 16th 2009, 07:57 AM
Jester
Quote:

Originally Posted by coolguy99
I'm getting... (x+3) / (5-x) as a sum right now... assuming that a = (x+3)/(2) and r = (x+3)/(2^n-1)

I got diverges, which I think is right, thanks :)

I'm still having trouble integrating..
u = lnx
du = 1/x dx
dv = 1/(x^p)
v = (p+1)/(x^p+1)

?

Write your dv as $dv = x^{-p}$ and take $p = 1$ as a special case.
• Mar 16th 2009, 08:04 AM
coolguy99
Alright, I see what you're getting at with that one... But how do you integrate vdu?

I split it up into ((x+3)/2) x ((x+3)/2))^(n-1)

I have the sum = (x+3) / (5-x), but I'm guessing that is incomplete?
• Mar 16th 2009, 10:45 AM
Jester
Quote:

Originally Posted by coolguy99
Alright, I see what you're getting at with that one... But how do you integrate vdu?

I split it up into ((x+3)/2) x ((x+3)/2))^(n-1)

I have the sum = (x+3) / (5-x), but I'm guessing that is incomplete?

Not sure how you got that sum $\frac{\frac{x+3}{2}}{1 - \frac{x+3}{2}} = - \frac{x+3}{x+1}$. As for your integration question

$u = \ln x,\; dv = x^{-p}dx$ or
$du = \frac{dx}{x},\; v =\frac{x^{1-p}}{1-p}$

so
$\int vdu = uv - \int u dv = \frac{x^{1-p} \ln x }{1-p} - \frac{1}{1-p} \int x^{1-p}\frac{dx}{x} = \frac{x^{1-p} \ln x }{1-p} - \frac{1}{1-p} \int x^{-p}\,dx$ $= \frac{x^{1-p} \ln x }{1-p} - \frac{1}{(1-p)^2} x^{1-p}$. Again, $p = 1$ is a special case.
• Mar 16th 2009, 07:57 PM
coolguy99
Alright, so that makes sense..

The question originally asked at what value for P would make the series convergent.

so you do the lim (as x->infinity) of
$\frac{x^{1-p} \ln x }{1-p} - \frac{1}{(1-p)^2} x^{1-p}
$

on the interval... (1, infinity)?

But I'm not quite sure how you would interpret that