Question

Find the sum of the series 1+ln(2)+(ln 2)^2/2!+...+(ln 2)^n/n!+...?

Answer 1

Well since e^x=1+x+x^2/2!+...+x^n/n! then you're really doing

e^ln2 which just equals 2.

Answer 2

So for problems like these, you must first memorize the Maclaurin series, right?

Answer 3

I suppose so. Often times you can recognize a Maclaurin series based on how its derivatives behave.

For instance, e^x is it's own derivative right? Similarly, that polynomial

1+x+x^2/2!+...+x^n/n!+...

is also its own derivative.

Answer 4

Which makes perfect sense, since they're exactly equal!

Answer 5

Thank you!

Answer 6

One way to remember sine and cosine is by remembering that:

e^(ix)=cosx+i*sinx

So by having e^x memorized, simply plugging in (ix) for (x) will give you the series for cosine (which is all even polynomial terms, 1, x^2, x^4, etc...) and the i will make it alternating.