# Alternating Series

• May 11th 2006, 11:39 AM
AfterShock
Alternating Series
"Let S = summation((-1)^i*1/[i*ln(i)-15], i = 10...infinity)

a.) Determine whether S diverges, converges conditionally, or converges absolutely.
b.) If S converges (whether conditionally or absolutely), find upper and lower bounds on S. If it diverges, find N so that S_N is >= 10.
c.) Let T represent the associated series consisting of all positive terms. That is, if S = summation((-1)^i*c_i, i = 10...infinity), let
T = summation(c_i, i = 10...infinity). If S converges absolutely, find N so that T_N is within .001 of T. If S does not converge absolutely, find N so that T_N >= 10."

This is what I did:

Since it's an alternating series, I can apply the alternating series test (AST). If the limit of the positive portion of it = 0, the servies S converges. That is, if limit as i -> infinity of 1/[i*ln(i) - 15] = 0, which it does. In order to determine whether it converges conditionally or absolutely, I need to take the absolute value of the series and determine if that converges or diverges.

Ok, if the absolute value of the series converges, then the series S converges absolutely. If it does not, that is, if it diverges, then the series S converges conditionally. To determine if the absolute value of the series converges/diverges, I need to use a positive series test. I decided to use a comparison test.

I compared 1/[i*ln(i) - 15] to 1/[i*ln(i)]:

1/[i*ln(i)] < 1/[i*ln(i) - 15]. Now I can do an integral test on 1/[i*ln(i)]...when doing that all out (im not going to list what I did because it'd take too long), we find out that 1/[i*ln(i)] diverges. Therefore, 1/[i*ln(i) - 15] must diverge too. Thus, it converges conditionally.

b.) I have to now find upper and lower bounds on S. Since S is an alternating series, as said before, I know S is between and two consecutive partial sums. I did S_10 and S_11. I concluded .0367 <= S <= .1246.

c.) I am stuck on this part. I don't really know what it's asking.

"c.) Let T represent the associated series consisting of all positive terms. That is, if S = summation((-1)^i*c_i, i = 10...infinity), let
T = summation(c_i, i = 10...infinity). If S converges absolutely, find N so that T_N is within .001 of T. If S does not converge absolutely, find N so that T_N >= 10."

Obviously, now that I know it converges conditionally, I have to find an N so that T_N >= 10.

Help would be greatly appreciated, as always.
• May 12th 2006, 09:01 AM
CaptainBlack
You will need to check this carefully!

Last part:

$\displaystyle T_N=\sum_{n=10}^{N} \frac{1}{n\log(n)-15}$

Now:

$\displaystyle n\log(n)-15<n\log(n)$

So:

$\displaystyle T_N>\sum_{n=10}^{N} \frac{1}{n\log(n)}>\int_{9}^{N-1}\frac{1}{x\log(x)}dx=\log \left( \frac{\log(N-1)}{\log(9)} \right)$

So now you need only find $\displaystyle N$ such that:

$\displaystyle \log \left(\frac{\log(N-1)}{\log(9)}\right) >10$

Which suggests that $\displaystyle N > e^{\log(9)e^{10}}+1$ will suffice

RonL
• May 12th 2006, 09:08 AM
AfterShock
log? or did you mean ln?
• May 12th 2006, 09:13 AM
CaptainBlack
Quote:

Originally Posted by AfterShock
log? or did you mean ln?

Default meaning of log is natural log, a text will make it clear that
it means common log if that is the default meaning in the text.

So where I have writen log I mean natural log, so you can replace
log by ln throughout what I have writen if that is the convention
you are used to.

RonL

PS

$\displaystyle e^{\log(9)e^{10}} \approx 3.89 \times 10^{21018}$

Impressive!!
• May 12th 2006, 11:43 AM
Jameson
In American schools we learn log to mean base 10 and ln to mean base e. For what reason we do this I don't know. Most higher level math literature I've read always uses log to mean base of e.
• May 12th 2006, 11:57 AM
CaptainBlack
Quote:

Originally Posted by Jameson
In American schools we learn log to mean base 10 and ln to mean base e. For what reason we do this I don't know. Most higher level math literature I've read always uses log to mean base of e.

and almost every programming language have log being the natural log

RonL
• May 16th 2006, 08:24 AM
topsquark
Quote:

Originally Posted by Jameson
In American schools we learn log to mean base 10 and ln to mean base e. For what reason we do this I don't know. Most higher level math literature I've read always uses log to mean base of e.

Thank you for mentioning that, Jameson, because whenever I read "log" I automatically think $\displaystyle log_{10}$ and it drives me crazy. (Well, crazier than usual. :) ) I almost always have to rewrite the expression so I can concentrate on the equations!

-Dan