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The sum to infinity of the series 1/1 + (1/1+2) + (1/1+2+3)+...... is equal to...

The sum to infinity of the series 1/1 + (1/1+2) + (1/1+2+3)+...... is equal to...

Grade:11

3 Answers

Meet
137 Points
6 years ago
The general term of the series is 1/[r(r+1)/2]= 2/[r(r+1)]. It can be written as 2[(1/r)-(1/r+1)], so for calculating sum put n in place of r therefore the answer is [2n/(n+1)]
Arun
25750 Points
6 years ago
Ww can write ggeneral term Tn 
Tn = 1/[n(n+1)/2]
Tn = 2/n(n+1)
We can simplify it by partial fraction
Tn = 2[1/n - 1/(n+1)]
Now we have to calculate the sum to infinite terms
T1 +T2 +T3+ T4+.......
= 2[1/1 - 1/2] + 2[1/2 - 1/3] +.....so on
Hence every terms cancel except the first terms
Hence sum  = 2 *1/1 = 2
Hope it helps
ankit singh
askIITians Faculty 614 Points
3 years ago

t’s actually simple….

Let series S= 1/1+1/3+1/6+1/10+1/15+…….

S can be rewritten as

S = 2 * (1/2+1/6+1/12+1/20+1/30+……..)

Now, let’s consider each term

1/2 = 1 - 1/2

1/6 = 1/2 - 1/3

1/12 = 1/3 - 1/4

1/20 = 1/4 - 1/5

1/30 = 1/5 - 1/6

Thus,

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