<p>What is the value of summation (from 1 to infinity) (-1)^n / n?</p>
<p>The solution said ln(1/2).</p>
<p>I have no idea how to work this out.
Please help.</p>
<p>It is a question come from 2002 FR Form B.
Anyone can help??</p>
<p>I have no idea. and I'm taking BC on Wednesday too! ahhhhhhhhh</p>
<p>anyone know a good review book that explains series well?</p>
<p>the barrons book is horrible.</p>
<p>thx</p>
<p>princeton review sucks too...</p>
<p>i dunno either, i tried the absolute value series and then the ratio test, but ended up with -1. i must be doing something wrong...</p>
<p>I think that is one of the special cases where a limit goes where it doesn't appear to go.
Either memorize it or use your calculator (Table function) to see where it approximates to.</p>
<p>that's the alternative harmonic...i say just memorize it :)</p>
<p>The solution said ln(1/2). = - ln(2) ok so if you have ln(1+x) the derivative is 1/(1+x) which transforms to 1/(1-(-x) which is a basic series formula that translates into 1 + u + u^2 + u^3 and then when you integrate it because that is the derivative on ln(1+u) you get
u + u^2/2 + u^3/3 which when you plug in -x for u you get -1 + 1/2 -1/3 which is the series for ln(2) but just negative and its formula is(-1)^n / n. And you have to know that its the series for ln(2) which is .693</p>
<p>I made a capture on the question and solution provided by my teacher...
From 2002 FR (Form B).</p>
<p>See whether anyone can help.</p>
<p>...like i said, just memorize it.</p>
<p>lol jk, thanx 4 the explanation matt :)</p>
<p>you need to realize that the Maclaurin series for ln(1/(1-x)) the infinite series from n = 1 to infinity is x^n/n. This yields, for x = -1, ln(1/(1-(-1)) = ln(1/(1+1)) = ln(1/2)</p>
<p>hey dude, thanks a lot.
I better put my head back to the defined Maclaurin series</p>
<p>Most review books don't seem to cover series too well. I have REA. which absolutely sucks, imo. and I borrowed Barron's today from the library- it was more helpful</p>
<p>u + u^2/2 + u^3/3 that is not really correct, you have to plug what u is and then integrate. Its how you derive the taylor series for tan-1 so if u = x^2 it would be x - x^3/3 + x^5/5 - x^7/7</p>
<p>In a practice test, I left all the maclaurin/taylor/power series stuff blank and I still got a strong 5.
I'll probably relearn it tomorrow or something...</p>