Question

(2) *(%) lim Eraluate KEO

Answer 1

First we evaluate
n k-1 3 -1 )k-1/n k 4
. Let
p = 4
, then the sum is

-1)k-1/n k p k k 0

The above sum can be written as (note the binomial expansion)

d pk Σ-11(ρ) kp1Σ-1) | ρ n k dp Κ k 0 k 0 η η Σ-11ρ )p kpk k, k dp k=0 k 0 η η Σ-1y-1! ) kp- k Σ-1()k dp k=0 k 0 η d Σ-11(p) kp-1- (1 -p)" k dp k=0 η Σ-11(μ kp-1-n1- P)"-1 κ k-0

Thus, setting ,

p = 4

k-1 n n-1 3 3 n1- -1)-k 4 4 k 0 k-1 3 n -11k 4 4n k 0

The limit,

k-1 n 3 k 4 n Σ-1 1 lim noo k 0 = 4 lim 4n 4 lim en In 4 k noo noo

You can use L'Hospital's rule.

.

3 1 lim-1 k - 4 lim In 4en in 4 4 noo noo k 0 k-1 3 n μm Σ-1)-1()k 0 n+o0 k 0 4