# Series and Ln

**Topics:**Series, Taylor series, Integral

**Pages:**3 (355 words)

**Published:**May 14, 2013

Instructions. Show all your work. Cell phones are strictly forbidden. Exam Duration : 70 min. 1. Show that 1 p n (ln n) n=2 converges if and only if p > 1: Solution: Apply integral test: Z Z ln R 1 X

R

2

1 p dx x (ln x) p=1 p 6= 1

let ln (x) = u then

ln 2

so that when p = 1 and p < 1 integral diverges by letting R ! 1, so does the series. When p > 1 then integral converges to ! 1 p 1 p 1 p (ln R) (ln 2) (ln 2) lim = , R!1 1 p 1 p 1 p so does the series. 2. (18 pts.) Find the in…nite sum 1 : n (n + 2) n=1 Solution: See that 1 1 = n (n + 2) n 1 n+2 1 X

8 R < ln ujln 2 ln 1 ln R du = 1 p : u p up 1

ln 2

hence 1 n (n + 2) n=1

k X

= =

n=1

1 1 1 1 1 + + + :::: 3 2 4 3 5 1 1 1 1 1 + + + k 2 k k 1 k+1 k 1 1 1 = 1+ + 2 k+1 k+2 1 ! 3 2

k X

1 n

1 n+2

1 k+2

so that

k X 1 1 = lim n (n + 2) k!1 n=1 n (n + 2) n=1

1 X

= lim

k!1

1 1 + k+1 k+2

=

3 2

1

3. (18 pts.) Find the Taylor series for f (x) = ln x at x = 4. Determine its interval of convergence. Solution: Recall that 1 X n

tn

= = =

1 1 t

;

n=0 1 X

jtj < 1 jtj < 1 jtj < 1

( 1) tn

n=0 1 X ( 1)n tn+1 n+1 n=0

1 ; 1+t

ln (1 + t) ;

let x

4 = t then ln (x) = = ln (4 + t) = ln 4 + ln 1 + ln 4 + t 4 = ln 4 + jx 1 X ( 1)n t n+1 4 ; n+1 n=0

t

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