# Math Help - Alternating Series

1. ## 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.

2. You will need to check this carefully!

Last part:

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

Now:

$
n\log(n)-15$

So:

$
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 $N$ such that:

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

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

RonL

3. log? or did you mean ln?

4. 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

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

Impressive!!

5. 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.

6. 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

7. 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 $log_{10}$ 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