infinite series

**muddywaters** · December 13th 2012, 02:13 AM

what's the difference between these two series?

1. $1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1}+\frac{1}{n+1}-...$

2. $(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+...+(\frac{1}{n}-\frac{1}{n+1})+...$

i think i understand how the 2nd one goes , but i don't get the first one. what difference does that last term make? and like in the 2nd one, the $(\frac{1}{n}-\frac{1}{n+1})$ towards the end gives the general term. but $\frac{1}{n}-\frac{1}{n+1}+\frac{1}{n+1}$ doesn't seem like the general term for the 1st series.

**HallsofIvy** · December 13th 2012, 07:58 AM

The "general term" is $a_n= \frac{1}{n/2}= \frac{2}{n}$ when when n is even, and $(-1)\frac{2}{n-1}$ when n is odd.

**muddywaters** · December 14th 2012, 09:00 PM

Originally Posted by HallsofIvy

The "general term" is $a_n= \frac{1}{n/2}= \frac{2}{n}$ when when n is even, and $(-1)\frac{2}{n-1}$ when n is odd.

i still do not understand. how did you find $a_n= \frac{1}{n/2}= \frac{2}{n}$ and $(-1)\frac{2}{n-1}$ ?
if that is the general term, then when $n=1$ ,then $(-1)\frac{2}{n-1} = (-1)\frac{2}{0}$ ? the first term when n=1 is should be $1$

maybe i should show how my book finds the sum the 2 series.

for the 2., it's straightforward enough.
$S_n=(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+...+(\frac{1}{n}-\frac{1}{n+1})=1-\frac{1}{n+1}$
$\lim_ {n \to \infty} S_n=1$
therefore the series converges and $S=1$

for 1., it's different, and i wonder why they do not use the same method to get the sum and how exactly are the 2 series different.
$S_2_m_-_1=1+(-\frac{1}{2}+\frac{1}{2})+(-\frac{1}{3}+\frac{1}{3})+...+(-\frac{1}{m}+\frac{1}{m})=1$
$S_2_m=1+(-\frac{1}{2}+\frac{1}{2})+(-\frac{1}{3}+\frac{1}{3})+...+(-\frac{1}{m}+\frac{1}{m})-\frac{1}{m+1}=1-\frac{1}{m+1}$
$\lim_ {n \to \infty}S_2_m_-_1=\lim_ {n \to \infty}S_2_m=1$
therefore, the series converges and S=1

**muddywaters** · December 19th 2012, 06:36 PM

i think i get it now.

for $(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+...+(\frac{1}{n}-\frac{1}{n+1})+...$ ,
the first term is $(1-\frac{1}{2})$ , second term is $(\frac{1}{2}-\frac{1}{3})$ , and so on
sum would be $1-\frac{1}{n+1}$

for $1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1}+\frac{1}{n+1}-...$ ,
the first term is $1$ , second term is $-\frac{1}{2}$ , third term is $\frac{1}{2}$ and so on
as it approaches infinity, there could be and odd number of terms or an even number of terms.
it isnt like the other series where every term come with pair of numbers and the numbers in the series surely consistently cancel each other out.
if there's an even number of terms, it would be like the other series, but if there's an odd number of terms, there will be less or more 1 term in comparison with the other series.
$S_2_m_-_1$ shows how the series would look if there's an odd number of terms and $S_2_m$ shows how it would be if there's an even number of terms

i could be wrong or inaccurate, this is just how i understand it

Thread: infinite series

LinkBack

Thread Tools

Display

infinite series

Re: infinite series

Re: infinite series

Re: infinite series

Similar Math Help Forum Discussions

[SOLVED] Fourier series to calculate an infinite series

Sum of an infinite series - comparing with a known series

Does convergence of an infinite series imply convergence of the infinite sequence?

Euler-Maclaurin Series formula on an infinite series?

calculus II areas about axis, pressure, infinite series, MacLaurin series

Search Tags