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