Theory and ExamplesEuler’s constant Graphs like those in

Chapter 10, Problem 57E

(choose chapter or problem)

Euler’s constant Graphs like those in Figure 10.11 suggest that as \(n\) increases there is little change in the difference between the sum

                

                                                          \(1+\frac{1}{2}+\cdots+\frac{1}{n}\)

and the integral

                                                           \(\ln n=\int_{1}^{n} \frac{1}{x} d x\)

To explore this idea, carry out the following steps.

a. By taking \(f(x)=1 / x\) in the proof of Theorem 9, show that

                                            \(\ln (n+1) \leq 1+\frac{1}{2}+\cdots+\frac{1}{n} \leq 1+\ln n\)

or

                                          \(0<\ln (n+1)-\ln n \leq 1+\frac{1}{2}+\cdots+\frac{1}{n}-\ln n \leq 1\)

Thus, the sequence

                                               \(a_{n}=1+\frac{1}{2}+\cdots+\frac{1}{n}-\ln n\)

is bounded from below and from above.

b. Show that

                                                \(\frac{1}{n+1}<\int_{n}^{n+1} \frac{1}{x} d x=\ln (n+1)-\ln n\)

and use this result to show that the sequence \(\left\{a_{n}\right\}\) in part (a) is decreasing.

Since a decreasing sequence that is bounded from below converges, the numbers \(a_{n}\) defined in part (a) converge:

                                                 \(1+\frac{1}{2}+\cdots+\frac{1}{n}-\ln n \rightarrow \gamma\)

The number \(\gamma\) whose value is 0.5772 . . ., is called Euler’s constant.

Equation Transcription:

 

Text Transcription:

n

1+1/2+cdots+1/n

ln n= integral over 1 ^n 1/x dx

f(x)=1/x

ln(n+1)leq 1+1/2+cdots+ 1/n leq 1+ln n

0<ln(n+1)-ln n1+1/2+cdots+1/n-ln n leq 1

a_n=1+1/2+cdots+1/n-ln n

1/n + 1 < integral over n ^n+1 1/x dx=ln (n+1)-ln n

{a_n}

a_n

1+1/2+cdots+1/n-ln n rightarrow gamma

gamma