What is the series for $\frac{e^x -1}{x}$?

Question

There is a question in my Calc review packet that says the following:

29. Use series to approximate the integral of $(e^x − 1)/x$ from $0$ to $1$. (Write out $4$ terms but do not add them.)

The answer given is: $S4 = 1 + 1/4 + 1/24 + 1/96$

How would you write the series for $(e^x - 1)/x$?

Answer 1

Basic approach. Take the following steps:

• Write out the Taylor series for $e^x$
• Subtract $1$ from the previous result
• Divide the previous result by $x$
• Take just the first four terms of the above, yielding a cubic polynomial
• Integrate this polynomial from $x = 0$ to $1$

Answer 2

$$e^x=1+x+\dfrac{x^2}2+\dfrac{x^3}{6}+\dfrac{x^4}{24}+\dots$$

$$e^x-1=x+\dfrac{x^2}2+\dfrac{x^3}{6}+\dfrac{x^4}{24}+\dots$$

$$\dfrac{e^x-1}x=1+\dfrac x2+\dfrac{x^2}6+\dfrac{x^3}{24}+\dots$$

$$\int _0^1\dfrac{e^x-1}x dx=x+\dfrac {x^2}4+\dfrac{x^3}{\color{red}{18}}+\dfrac{x^4}{96}+\dots\large\mid_0^1$$

$$=1+\dfrac14+\dfrac1{\color{red}{18}}+\dfrac1{96}+\dots$$

There must have been a mistake in the answer key.

Answer 3

$(e^t-1)/t=1/(\sum_{k\ge0}B_m^-t^m/m!)$, where the $B_m^-$ are Bernoulli numbers.