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Mathematics Dictionary
Dr. K. G. Shih

Special series

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Q01 | - S = 1/(1*2) + 1/(2*3) + ..... + 1/(n*(n+1))

Q02 | - S = 1/(3*5) + 1/(5*7) + ...... + 1/((2*n-1)*(2*n+1))

Q03 | - S = 1/(1*2*3) + 1/(2*3*4) + 1/(3*4*5) + ...... + 1/(n*(n+1)*(n+2)

Q04 | - S = (1-1/4)*(1-1/9)*(1-1/16)*.....*(1-1/(n^2)) where n GT 2

Q05 | -

Q06 | -

Q07 | -

Q08 | -

Q09 | -

Q10 | -

Answers

Q01. S = 1/(1*2) + 1/(2*3) + ..... + 1/(n*(n+1))

By simple observation

1/(1*2) = 1 - 1/2
1/(2*3) = 1/2 - 1/3
1/(3*4) = 1/3 - 1/4
1/(n*(n+1) = 1/n - 1/(n+1)
Sum above terms we have S = 1 - 1/(n+1)

By induction : 1/(1*2) + 1/(2*3) + ..... + 1/(n*(n+1)) = 1 - 1/(n+1)

n =1, L.S. = 1/2 and R.S. = 1/2. Hence L.S. = R.S. is true.
n =2, L.S. = 1/2+1/6 = 2/3 and R.S. = 1-1/3 = 2/3. Hence L.S. = R.S. is true.
n =3, L.S. = 1/2+1/6+1/12 =3/4 and R.S. = 1-1/4 = 3/4. Hence L.S. = R.S. is true.
L.S. = S + 1/((n+1)*(n+2).
R.S. = 1 - 1/(n+1) + 1/((n+1)*(n+2).
R.S. = 1 - ((n+2)-1)/((n+1)*(n+2)).
R.S. = 1 - (n+1)/((n+1)*(n+2)).
R.S. = 1 - 1/((n+1)+1). Hence (n+1)th term give same result as nth term.

Q02. S(n) = 1/(1*3) + 1 /(3*5) + 1/(5*7) + ...... + 1/((2*n-1)*(2*n+1))

Method 1 : By simple observation

1/(1*3) = (1 - 1/3)/2
1/(3*5) = (1/3 - 1/5)/2
1/(5*7) = (1/5 - 1/7)/2
1/((2*n-1)*(2n+1)) =(1/(2*n-1) - 1/(2*n+1))/2
Sum above terms we have S = (1 - 1/(2*n+1))/2 = n/(2n+1)

Method 2 : By induction : 1/(1*3) + 1/(3*5) + ..... + 1/((2*n-1)*(2*n+1)) = (1 - 1/(2*n+1))/2

n =1, L.S. = 1/3 and R.S. = 1/3. Hence L.S. = R.S. is true.
n =2, L.S. = R.S. = 2/5. Hence L.S. = R.S. is true.
n =3, L.S. = R.S. = 3/7. Hence L.S. = R.S. is true.
L.S. = S + 1/((2*n+1)*(2*n+3). We have (n+1) terms.
R.S. = (1 - 1/(2*n+1))/2 + 1/((2*n+1)*(2*n+3).
R.S. = 1/2 - ((2*n+3)-2)/(2*(2*n+1)*(2*n+3)).
R.S. = 1/2 - (2*n+1))/(2(2*n+1)*(2*n+3)).
R.S. = (1 - 1/((2*n+3). Hence (n+1)th term give same result as nth term.

Method 3 : By finding S(1), S(2), ...

S(1) = 1/3
S(2) = 1/3 + 1/15 = 2/5
S(3) = 1/3 + 1/15 + 1/35 = 2/5 + 1/35 = 3/7
S(4) = 1/3 + 1/15 + 1/35 + 1/63 = 3/7 + 1/63 = 4/9
By observation : S(n) = n/(2*n+1)

Q03. S = 1/(1*2*3) + 1/(2*3*4) + 1/(3*4*5) + ...... + 1/(n*(n+1)*(n+2)

1/(1*2*3) = (4/(2*3)/4 = (1 - 2/(2*3))/4
1/(2*3*4) = (4/(2*3*4)/4 = (2/(2*3) - 2/(3*4))/4
1/(3*4*5) = (4/(3*4*5)/4 = (2/(3*4) - 2/(4*5))/4
Hence S = (1 - 2/((n+1)*(n+2))/4

Induction method

both sides add 1/((n+1)*(n+2)*(n+3)
Prove R.S. = (1 - 2/(n+2)*(n+3))/4

Q04. S = (1-1/4)*(1-1/9)*(1-1/16)*.....*(1-1/(n^2)) where n is greater than 2

n = 2, S = 3/4 ............................ (2+1)/(2*2)
n = 3, S = (3/4)*(8/9) = 2/3 = 4/6 ........ (3+1)/(2*3)
n = 4, S = (3/4)*(2/3)*(15/16) = 5/8 = .... (4+1)/(2*4)
By observation, we have
n = k, S = (k+1)/(2*k)

Induction method

Both side times (1 - 1/((n+1)^2)
Hence R.S. = ((n+1)/(2*n))*(1-1/((n+1)^2)
R.S. = (n+2)/(2*(n+1)). That is sum of n term and (n+1) terms have same form.

Find S if n goes to infinite

Lim[(n+1)/(2*n)]
= Lim[(1+1/n)/2] = 1/2 as n goes to infinite

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