I'm a little confused as to what they are asking. all the examples of taylor series expansion I have seen use x instead and I'm not sure how I would go expanding these series.
$\begingroup$
$\endgroup$
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
Hints:
You probably know or can find the Taylor expansion for $e^x$. Call this $f(x)$.
Now consider the derivative of $\dfrac{f(x)-1}{x}$ and see what happens what happens to the coefficients of the powers of $x$. Then let $x=1$ and consider what series this gives. Then consider the derivative of $\dfrac{e^x-1}{x}$ when $x=1$.
Can you do something similar to get the second series? You might consider integration or the answer to the first series.
$\begingroup$
$\endgroup$
The expansion for $e^x$ is $$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots$$ Notice that $$\frac{e^x-1}{x}=1+\frac{x}{2!}+\frac{x^2}{3!}+\dots$$ And if you differentiate it, $$\frac{xe^x-(e^x-1)}{x^2}=\frac{1}{2!}+\frac{2x}{3!}+\frac{3x^2}{4!}+\dots$$
Now, you only need to substitute $x=1$ to get the first summation.
I think you need to do something similar for the second series.