Infinite series solution

[median kiln]

I can't even simplify this correctly to get the solution I need to learn how to sum this infinite series

[grizzled plinth]

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[burnt bloom]

First of all you can try re-indexing that

and/or add terms conveniently so that you get something you recognize

here's another bone

$$S(x) = \sum_{n=1}^{\infty} \frac{1}{(n+1)!} x^{n-1}$$

[sly sequoia]

Rion

[burnt bloom]

is S differentiable? if yes, what is S'?

[median kiln]

this is the solution

but I can't figure out what is going on here

why is this simplied as is

we canceled out the n+1 but why is there second sum out of nowhere

[burnt bloom]

expanding (n+1-2) = (n+1) - 2

[median kiln]

thats right

I understand

that

but the second sum with the original denominator is confusing me hell so much

[burnt bloom]

$\sum (a_n + b_n) = \sum a_n + \sum b_n$

[sly sequoia]

Rion

[burnt bloom]

no?

$a_n = \frac{(-1)^n (n+1)}{(n+1)!}, b_n = -\frac{2 (-1)^n}{(n+1)!}$

[sly sequoia]

Rion

[median kiln]

can be n! written as (n+1)n! and (n+1)n! as just n!?

[burnt bloom]

(n+1)! = (n+1) * n!

I don't understand your question

[median kiln]

when we cancel out the (n+1) then denominator is just n! in the An

how can we then get n+1 back in the Bn?

no?

[burnt bloom]

no

you split it into two terms first

$\sum_{n=1}^{\infty} \left( \frac{(-1)^n (n+1)}{(n+1)!} -\frac{2 (-1)^n}{(n+1)!}\right)$

[sly sequoia]

Rion

[burnt bloom]

$= \sum_{n=1}^{\infty} \frac{(-1)^n (n+1)}{(n+1)!} - \sum_{n=1}^{\infty} \frac{2 (-1)^n}{(n+1)!}$

[sly sequoia]

Rion

[median kiln]

OMGG

its like we transformed the n-1 in to ((n+1)-2) so it's n+1 * -(-1)^n AND -2 * (-1)^n?

[burnt bloom]

yes, that's what i've been trying to say

[median kiln]

OMG

I LOVEE YOUU

more than my girlfriend probably

jk jk

but I really love youu

almost 6 hours of my lifee

[umbral inlet]

@median kiln

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