Question

Oh I think I get the idea of stripping out a term now.
Simply by taking out an \[a_1\] changes where the sum starts? Yes?
\[\sum_{n=1}^\infty a_n=a_1+\sum_{n=2}^{\infty} a_n\]

Answer 1

so when I have \[\sum_{n=1}^{\infty}a_nnx^n=a_1+\sum_{n=2}^{\infty}a_nnx^n\]
but it's not n=0? Why?

Answer 2

I'm trying to start at n=0

Answer 3

\[\sum_{n=1}^{\infty}a_nnx^n \rightarrow \sum_{n=0}^{\infty}a_nnx^n\]

Answer 4

I looked at example 1 a.
http://tutorial.math.lamar.edu/Classes/CalcII/Series_Basics.aspx
but I'm not trying to do an index shift :C

Answer 5

Well do you need to start at 0? because if you want n = 0 you can just make it at the bottom n = 0

Answer 6

why though?

Answer 7

why is that all I need to do?

Answer 8

\[\sum_{n =0}^{\infty}\]

Answer 9

because in your equation it would work for not getting an answer anyway

Answer 10

because multiplying by 0 just leaves the term out basically, so you can have it as n=0 but you won't get an actual term out.

Answer 11

if you are trying to get a term out there have to be different conditions met in your equation otherwise it won't work that way. But it looks fine for getting a term out if it's not n = 0.

Answer 12

by that do you mean that
\[\sum_{n=1}^{\infty}a_nnx^n=a_1+\sum_{n=0}^{\infty}a_nnx^n\]
It leaves the \[a_1\]out?

Answer 13

no

Answer 14

:(

Answer 15

it's right the left part

Answer 16

but if you made that n = 0

Answer 17

on the left and n= 1 on the right

Answer 18

it would be \[\alpha _{0}*0*x^0\]

Answer 19

OMG! I think something just clicked!

Answer 20

Yeah, you see where im going with it

Answer 21

no wait, just a second

Answer 22

You can have it start at 0.. but it doesnt change anything if you start at 1 in this case

Answer 23

\[\sum_{n=0}^{\infty}a_nnx^n=0\]

Answer 24

it's like putting 0 + \[\sum_{n=1}^{\infty} (insert rest here)\]

Answer 25

for n = 0

Answer 26

I guess if you want to start at 0, it has to be fixed at 0 and you can't change that value for it so it's usually left out of summations and n starts at 1. There are situations where you can have n start at 0 but in you equation those conditions for having a different term at n = 0 is not met.

Answer 27

I'm a very visual learner.
Where you said: "on the left and n= 1 on the right"
\[\sum_{n=1}^{\infty}a_nnx^n=a_1+\sum_{n=0}^{\infty}a_nnx^n\]
^^^Is wrong I know, but would you mind writing what you mean?

Answer 28

I'm honestly trying hard to get this...I think we're almost there...sorry =/

Answer 29

\[\sum_{n=0}^{\infty}a _{n}nx ^{n} = 0 + \sum_{n=1}^{\infty}a _{n}nx ^{n}\]

Answer 30

using the rule that 0 multiplied by anything is 0

Answer 31

Thank you, finally!

Answer 32

Lol it's ok, glad i could help