What is the formula for $\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\cdots +\frac{1}{n(n+1)}$

Question

How can I find the formula for the following equation?

$$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\cdots +\frac{1}{n(n+1)}$$

More importantly, how would you approach finding the formula? I have found that every time, the denominator number seems to go up by $n+2$, but that's about as far as I have been able to get:

$\frac12 + \frac16 + \frac1{12} + \frac1{20} + \frac1{30}...$ the denominator increases by $4,6,8,10,12,\ldots$ etc.

So how should I approach finding the formula? Thanks!

score 24 · Answer 1 · answered Jan 24 '13 at 18:20

24

Hint: Use the fact that $\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$ and find $S_n=\sum_1^n\left(\frac{1}{k}-\frac{1}{k+1}\right)$.

answered Jan 24 '13 at 18:20

Mikasa

67,374

Simple and nice (+1) – user 1591719 Jan 24 '13 at 18:59
@Chris'ssister: Thanks a lot for your consideration. – Mikasa Jan 24 '13 at 19:03
Nicely said, Babak(+1)! – amWhy Jan 24 '13 at 19:21

score 13 · Accepted Answer · answered Jan 24 '13 at 18:21

13

If you simplify your partial sums, you get $\frac12,\frac23,\frac34,\frac45,....$ Does this give you any ideas?

answered Jan 24 '13 at 18:21

Cameron Buie

102,994

Thanks! This was very helpful :) Now I just gotta prove it by induction! – Charles Jan 24 '13 at 18:31
I dont get it... how that can be – Dini Oct 01 '19 at 20:03
@Lifeforbetter Can you be more specific about what you find confusing? – Cameron Buie Oct 01 '19 at 20:15
How 1/6 become 2/3 1/12 become 3/4 – Dini Oct 02 '19 at 05:20
@Lifeforbetter Oh, my! That would be confusing! Rather, $$\frac12+\frac16=\frac46=\frac23,$$ $$\frac23+\frac1{12}=\frac9{12}=\frac34,$$ and so on. – Cameron Buie Oct 02 '19 at 09:42
So n/(n+1) is the sum? – Dini Oct 02 '19 at 14:47
@Lifeforbetter Precisely! – Cameron Buie Oct 02 '19 at 18:47
Note that $\dfrac{k}{k+1}+\dfrac{1}{(k+1)(k+2)}=\dfrac{k(k+2)+1}{(k+1)(k+2)}=\dfrac{(k+1)^2}{(k+1)(k+2)}=\dfrac{k+1}{k+2}$ – Henry Jul 13 '23 at 09:09

score 3 · Answer 3 · edited Oct 01 '17 at 22:47

3

While exploitation of the resulting telescoping series after partial fraction expansion is a very simple way forward, I thought it might be instructive to present another way forward. Here, we write

$$\begin{align} \sum_{k=1}^N \frac{1}{k(k+1)}&=\sum_{k=1}^N \int_0^1y^{k-1}\,dy \int_0^1 x^k\,dx\\\\ &=\int_0^1\int_0^1x\sum_{k=1}^N (xy)^{k-1}\,dx\\\\ &=\int_0^1\int_0^1 x\frac{1-(xy)^N}{1-xy}\,dx\,dy\\\\ &=\int_0^1\int_0^x \frac{1-y^N}{1-y}\,dy\,dx\\\\ &=\int_0^1\int_y^1\frac{1-y^N}{1-y}\,dx\,dy\\\\ &=\int_0^1(1-y^N)\,dy\\\\ &=1-\frac1{N+1} \end{align}$$

as expected!

edited Oct 01 '17 at 22:47

Simply Beautiful Art

74,685

answered Mar 05 '16 at 22:14

Mark Viola

179,405

1

Where can i learn more about this technique ? Any references/books ? – Krosin May 12 '21 at 16:14

Harsh Kumar · Answer 4 · 2017-01-11T15:59:57.553

3

Observe that $$\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$$ $\therefore$ The given series can be written as $$1-\frac12+\frac12-\frac13+\frac13+\cdots -\frac{1}{n}+\frac1n-\frac{1}{n+1}$$ $$=1-\frac1{n+1}$$ $$=\frac{n}{n+1}$$

edited Jan 11 '17 at 15:59

answered Jan 09 '17 at 17:21

Harsh Kumar

2,846

score 1 · Answer 5 · answered Jun 03 '21 at 04:16

You can use these formulas for Problems like this , $$\frac{1}{a.b}=\frac{1}{(b-a)}\left(\frac{1}{a}-\frac{1}{b}\right)$$ $$\frac{1}{a.b.c}=\frac{1}{(c-a)}\left(\frac{1}{a.b}-\frac{1}{b.c}\right)$$

Hope it'll help you .

What is the formula for $\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\cdots +\frac{1}{n(n+1)}$

5 Answers5

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