How exactly can I tell a series with different index equals another series, could I just plug in numbers?

Question

astrophysics · Answer

For example I was working on a DE yesterday, but I was not exactly sure why this is true, $$\sum_{n=2}^{\infty} a_nn(n-1)x^{n} = \sum_{n=0}^{\infty} a_nn(n-1)x^{n} $$ it's more the a_n that confuses me.

astrophysics · Answer

@amistre64 @jim_thompson5910 @IrishBoy123

irishboy123 · Answer

astro

i personally would just plug them in.  i am not sure, however, that that is how you are "supposed" to do it.

hope this finds you well 😊

empty · Answer

You can see that everything is the same, except for two missing terms of n=0 and n=1, $$\sum_{n=2}^{\infty} a_nn(n-1)x^{n} = \sum_{n=0}^{\infty} a_nn(n-1)x^{n} $$ 

So to prove that, you can pull out the first 2 terms of the sum on the right:

$$\sum_{n=2}^{\infty} a_nn(n-1)x^{n} = a_0 0(0-1) + a_1(1(1-1))+ \sum_{n=2}^{\infty} a_nn(n-1)x^{n} $$ 
Now wait, these two sums are the same, so we can remove them by subtracting from both sides:

$$0= a_0 0(0-1) + a_1(1(1-1))$$

Ok, so now we're left with this, which is I think a pretty clearly true statement, so it's all true.

astrophysics · Answer

Oooooh I think I get it, so we can start at n = 0 because when you get to n=2 it's just the same sum, as the n= 0 and n = 1 go to 0 anyways, ahh I hope I said that correctly! Thanks man!

Thanks to @IrishBoy123 as well

empty · Answer

Yeah exactly, those two contributing terms are just 0, so adding 0 never changed nuffin' :P

astrophysics · Answer

That makes so much sense haha

empty · Answer

Don't say that, this is diffy Q it's not supposed to make sense r u outta ur mind?!

astrophysics · Answer

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anonymous · Answer

@AloneS