## anonymous 4 years ago 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? $\sum_{n=1}^\infty a_n=a_1+\sum_{n=2}^{\infty} a_n$

1. anonymous

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?

2. anonymous

I'm trying to start at n=0

3. anonymous

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

4. anonymous

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

5. anonymous

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

6. anonymous

why though?

7. anonymous

why is that all I need to do?

8. anonymous

$\sum_{n =0}^{\infty}$

9. anonymous

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

10. anonymous

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.

11. anonymous

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.

12. anonymous

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?

13. anonymous

no

14. anonymous

:(

15. anonymous

it's right the left part

16. anonymous

but if you made that n = 0

17. anonymous

on the left and n= 1 on the right

18. anonymous

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

19. anonymous

OMG! I think something just clicked!

20. anonymous

Yeah, you see where im going with it

21. anonymous

no wait, just a second

22. anonymous

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

23. anonymous

$\sum_{n=0}^{\infty}a_nnx^n=0$

24. anonymous

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

25. anonymous

for n = 0

26. anonymous

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.

27. anonymous

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?

28. anonymous

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

29. anonymous

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

30. anonymous

using the rule that 0 multiplied by anything is 0

31. anonymous

Thank you, finally!

32. anonymous

Lol it's ok, glad i could help