Find the sum of 1/((n)(n-1)) from n = 1 to 100.
write 1= n-(n-1)
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)
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).
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.
There is another solution based on the theory of inhomogeneous sequences.The sum (depending only by 'n') must be seeked in the form:
We know that:
Introducing (1) in (2) (3) & (4) and solving the system --->
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).
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