limit and convergent

Question

lim   (1/n+1 + 1/n+2 + ......... 1/2n) is equal to

^n→∞ 

a) 0 

b) 1

c) 2

d) log_e2

the answer is log_e2 , bt howi cant understand, it should be zero i think, please solve it, plz plz

 

Answer 1

Divide the interval [1, 2] into n equal subintervals.  Each subinterval has length 1/n.  Let f(x) = 1/x and form the following Riemann sum using right endpoints:

 

f(1 + 1/n)(1/n) + f(1 + 2/n)(1/n) + f(1 + 3/n)(1/n + ... + f(1 + n/n)(1/n)

 

Now, f(1 + 1/n) = 1/(1+1/n) = n/(n+1)

 

        f(1 + 2/n) = 1/(1 + 2/n) = n/(n+2)  etc

 

Denote the Riemann Sum by S_n.

 

S_n = (1/n)[n/(n+1) + n/(n+2) + ... + n/(2n)] 

 

    = 1/(n+1) + 1/(n+2) + 1/(n+3) + ... + 1/(2n)

 

lim S_n (as n → ∞) = ∫(1/x)dx (from x = 1 to x = 2)

 

                          = ln2 - ln1 = ln2 - 0 = ln2 

 

 

Answer 2

The series as you have written it diverges  (it is the harmonic series).

 

However a modification of this series converges conditionally to ln(2).   The modification is to consider the alternating sign version :

 

   1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 +  ...

 

   The Leibniz alternating series test shows that it converges conditionally.   A trick is needed to see that it converges to ln(2)

 

   The trick is to consider the Taylor series expansion of ln(y) about the point y =1 .   This Taylor series can be

written as

 

   ln( 1 +x)  = 1 - (1/2) x + (1/3) x² - (1/4) x³ - (1/5) x⁵ - (1/6) x⁶ + ...

 

  Plugging in x =1 into this form results in

 

  ln(2)  =  1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6  +  ...

 

 

  This is the desired result.   It is well known  to mathematicians, but is a bit of a curiosity piece.