Find the sum of 1/((n)(n-1)) from n = 1 to 100

[PrudensOptimus]

Find the sum of 1/((n)(n-1)) from n = 1 to 100.

Any idea?

 

[Hurkyl]

Partial fractions!

 

[himanshu121]

write 1= n-(n-1)

 

[HallsofIvy]

One "idea" I have is that the sum does not exist. The fraction is undefined when n= 1. Did you mean to start at n= 2 or would you prefer to use 1/n(n+1)? (which would be the same thing)

 

[metacristi]

Let SUM=∑_{k=2 to n} [1/(k(k-1)] (1)


Apart from the straightforward method using partial fractions the seeked sum can be found from a more general approach (as a particular solution).

Let

S=ln[(x+2)/(x+3)]+ln[(x+3)/(x+4)]+...+ln[(x+n)/(x+n+1)]

S=ln{[(x+2)/(x+3)]*[(x+3)/(x+4)]*..*[(x+n-1)/(x+n)]*[(x+n)/(x+n+1]}


S=∑_{k=2 to n} [ln(x+k)-ln(x+k+1)]=ln(x+2)-ln(x+n+1)


By deriving both members --->


S'=∑_{k=2 to n} [1/(x+k)(x+k+1)]=[(n-1)/(x+2)(x+n+1)]


If we put now x=-1 results:


SUM=∑ _{k=2 to n} [1/(k(k-1)]=(n-1)/n=1-(1/n) (2)


The adavantage of this method is clear,the general formula depending on x can be particularized to find other sums,∑ _{k=1 to n} [1/(k(k+1)] for example,and so on.

 

[metacristi]

There is another solution based on the theory of inhomogeneous sequences.The sum (depending only by 'n') must be seeked in the form:

SUM=(A/n²)+(B/n)+C (1)

A,B,C=constants

We know that:

S[n=2]=1/2 (2)

S[n=3]=2/3 (3)

S[n=4]=3/4 (4)

s[n=5]=4/5 (5)

S[n=6]=5/6 (6)

Introducing (1) in (2) (3) & (4) and solving the system --->

A=0
B=-1
C=1

Therefore SUM=(-1/n)+1=1-(1/n) (7)

Introducing now in (7) n=5 ---> S[n=5]=4/5 that is exactly (5) (for n=6 we obtain again an equality).The solution obtained is exact and holds for all values of 'n' (actually the inhomogeneous theory requires to prove that the sequence made with the free terms-here the general terms in 'n'=[1/(n(n-1))]-constitute a linear sequence but in practice it is enough to show that the obtained equation S=f(n) holds exactly for n greater than the values used to calculate the constants).