Question

Can someone help me with a series of problems concerning series? Will type up in comments.

Answer 1

1. Find the sum of the series
\[\sum_{n=3}^{\infty} \frac{ 1 }{ (2n-3)(2n-1) }\]

2.The series is the value of the Taylor series at x = 0 of a function f(x) at a particular point. What function and what point?
\[1 + \ln 2 + \frac{ (\ln 2)^{2} }{ 2! } + ... \frac{ (\ln 2)^{n} }{ n! } + ...\]

3. Find Taylor series at x=0 for the function.
\[\frac{ 1 }{ 1-2x }\]

Answer 2

@Best_Mathematician there will be some stuff like this on your AP test today
this can be a brain warm-up :)

Answer 3

a partial fraction decomp might be useful on the first one

Answer 4

the second one looks familiarly like \[e^x = \sum\frac1{n!}x^n\]

Answer 5

and a a simple method, if it can be employed, to find a power series, is just to do longhand division


1+2x+4x^2 .... \(=\sum2n~x^n\) maybe
------------
1-2x | 1
-(1-2x)
--------
2x
-(2x-4x^2)
----------
4x^2

Answer 6

2^n that is

Answer 7

@amistre64 I understand #3 now, but about #2, is that just something you have to know (and it should look familiar?)

Answer 8

that is something which you should remember yes. whenever you see an ln floating around there is bound to be an e^.... in the process

Answer 9

the derivative of e^x is just e^x ... and at x=0, thats just a bunch of 1s for constant

\[e^x=\sum\frac{f^{(n)}(0)}{n!}x^n\to\sum\frac{1}{n!}x^n\]

Answer 10

the interval of convergence is therefore\[|x|\lim\frac{(n)!}{(n+1)!}\]
\[|x|\lim\frac{1}{n}=0x\]
since 0x < 1 is true for any value of x, the interval of convergence is the set of Reals

Answer 11

1/(n+1) but not alot of difference from that

Answer 12

Oh, I see. Also, on an unrelated note, concerning interval of convergence, I was wondering if it was possible to write an interval of \(-3<x^{2}<3\) to be \(-\sqrt{3} <x<\sqrt{3}\) ? I don't see how it can be, but perhaps I need a little review in algebra...

Answer 13

thats fine, the middle there for quadratics can get tricky; but since |x^2| = x^2 its no big deal in this case

Answer 14

when we get into stuff like |x^2-5| < 3