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$\sum\limits_{x=1}^{99}\frac{1}{x(x+1)}$ = __________________.

edited | 2.5k views
+11
The summaration of series is

$\frac{1}{1.2}+\frac{1}{2.3}+ \dots +\frac{1}{99.100}\\=\frac{2-1}{1.2}+\frac{3-2}{2.3}+ \dots +\frac{100-99}{99.100}\\=\frac{2}{1.2} + \frac{-1}{1.2}+\frac{3}{2.3}+\frac{-2}{2.3}+ \dots +\frac{100}{99.100}+\frac{-99}{99.100}\\=1-\frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4}+ \dots +\frac{1}{99} - \frac{1}{100}\\=1-\frac{1}{100} = 0.99$

$\frac{1}{1.2}+\frac{1}{2.3}+ \dots +\frac{1}{99.100}\\=1-\frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4}+ \dots +\frac{1}{99} - \frac{1}{100}\\=1-\frac{1}{100} = 0.99$
by Veteran (431k points)
selected by
+1
Sorry for being so naive , but how did you get to step 2 ?

how did you derive it ?
+4
Ok got it fractional expansion , and Harmonic Series.

[0-99] 1/x - 1/(x+1)

Step 2 is just expansion.
+6
Yes, it is just expansion. But may be something more intuitive can be told. Let me think.
+6

Waiting for " something more intuitive " !

+3
For every term in the series, the 1 in the numerator can be written as the difference of the 2 terms in the denominator e.g. 2-1/1.2 + 3-2/2.3 + 4-3/4.3..... And this leads to the expansion given.
0
Summation 1-99 1/x(x+1) is  telescopic series.
It can be answered by using the concept of partial fraction

$\dfrac{1}{x\left(x+1\right)} =\dfrac{A}{x} + \dfrac{B}{\left(x+1\right)}$

solving this we will get $A=1$ and $B=-1$

So, this will form a sequence in which $2$ terms will remain $\left(1-\dfrac{1}{100}\right)$ and we will get $\dfrac{99}{100}=0.99$ as answer.
by Junior (993 points)
edited
0
Can u please elaborate to solve this.. I forgot now
+4
1/x(x+1) = A/x + B/(x+1)

1/x(x+1) = (A(x+1) + Bx)/x(x+1);   // take LCM

1/x(x+1) = (A(x+1)+Bx)/x(x+1); // cancel x(x+1) on RHS and LHS

then 1 = A(x+1) + Bx

1 = x(A + B) + A; // compare LHS and RHS on the basis of degree of x

A + B = 0 and A = 1

solve it .. therefore A = 1 and B = -1
If u take 2 instead of 99 u will get 2/3

If u take 3  u will get 3/4

Means behavior remains same...

So for 99 u will get 99/100
by Junior (739 points)
0.99
by Active (1.2k points)
$\sum_{x = 1}^{99} = 1/x *(1+x) = (1/x) - (1/1+x)$

$(1/x) - (1/1+x) = (1+x - x) / x*(x+1) = 1/x *(1+x)$

$s_{1} = 1 - 1/2$

$s_{2} = 1/2 - 1/3$

$s_{3} = 1/3 - 1/4$

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$s_{99} = 1/99 - 1/100$

$s_{1} + s_{2} + s_{3} + ...............................................+ s_{99}= \sum_{x = 1}^{99} = 1 - 1/100 = 99/100 = 0.99$