Thread: sum of 1st nth terms

1. sum of 1st nth terms

If the $r$th term of the series
$\frac{1}{1}+\frac{1}{1+2}+\frac{1}{1+2+3}+$... is $\frac{1}{1+2+3+...+r}$
show that the sum of the first $n$ terms of the series is $\frac{2n}{n+1}$.

2. Originally Posted by janvdl
You have the first and nth terms.

$S_{n} = n \left( \frac{a + l}{2} \right)$

Where a is the first term and l the last(or the nth term in this case).
This formula is foe A.P.
The series is A.P.?

3. Originally Posted by SengNee
The series is A.P.?
Of course it is not A.P.
Because $a_r = \frac{1}{{1 + 2 + \cdots + r}} = \left( {\sum\limits_{k = 1}^r k } \right)^{ - 1} = \frac{2}{{r\left( {r + 1} \right)}}$
we get $\sum\limits_{k = 1}^n {a_k } = 2\sum\limits_{k = 1}^r {\left( {\frac{1}{k} - \frac{1}{{k + 1}}} \right) = \frac{{2n}}{{n + 1}}}
$
.

4. $\frac{1}{1+2+\cdots + k} = \frac{2}{k(k+1)}$

This means you want to sum, $\sum_{k=1}^n \frac{2}{k(k+1)} = 2 \sum_{k=1}^n \frac{1}{k(k+1)}$.

Now, $\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$.

Thus, $2\sum_{k=1}^n \frac{1}{k} - \frac{1}{k+1} = 2\left( 1 - \frac{1}{n+1} \right) = \frac{2n}{n+1}$.

5. Originally Posted by ThePerfectHacker
$\frac{1}{1+2+\cdots + k} = \frac{2}{k(k+1)}$

This means you want to sum, $\sum_{k=1}^n \frac{2}{k(k+1)} = 2 \sum_{k=1}^n \frac{1}{k(k+1)}$.

Now, $\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$.

Thus, $2\sum_{k=1}^n \frac{1}{k} - \frac{1}{k+1} = 2\left( 1 - \frac{1}{n+1} \right) = \frac{2n}{n+1}$.
Why $\frac{1}{1+2+\cdots + k} = \frac{2}{k(k+1)}$ ?

6. Originally Posted by SengNee
Why $\frac{1}{1+2+\cdots + k} = \frac{2}{k(k+1)}$ ?
Because $1+\cdots + k = \frac{k(k+1)}{2}$ --> arithmetic series.

7. Hello, SengNee!

Given the series: . $S \;=\;\frac{1}{1}+\frac{1}{1+2}+\frac{1}{1+2+3}+ \hdots +\frac{1}{1+2+3+\hdots +n} + \hdots$

show that the sum of the first $n$ terms of the series is: . $S_n \;=\;\frac{2n}{n+1}$
This is neither an Arithmetic Series nor a Geometric Series.
. . We must find another approach . . .

The denominators are "triangular numbers" ... of the form: . $\frac{n(n+1)}{2}$

We have: . $S_n \;=\;\frac{2}{1\cdot2} + \frac{2}{2\cdot3} + \frac{2}{3\cdot4} + \frac{2}{4\cdot5} + \hdots + \frac{2}{n(n+1)}$

. . . . . . . . . . $=\;2\left[\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \hdots + \frac{1}{n(n+1)}\right]$

. . . . . . . . . . $= \;2\left[\left(\frac{1}{1}-\frac{1}{2}\right) + \left(\frac{1}{2}-\frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \hdots + \left(\frac{1}{n} - \frac{1}{n+1}\right)\right]$

All the "middle" terms cancel out and we are left with:

. . . . . $S_n \;\;=\;\;2\left[1 - \frac{1}{n+1}\right] \;\;=\;\;\frac{2n}{n+1}$

8. Originally Posted by Soroban
Hello, SengNee!

This is neither an Arithmetic Series nor a Geometric Series.
. . We must find another approach . . .

The denominators are "triangular numbers" ... of the form: . $\frac{n(n+1)}{2}$

We have: . $S_n \;=\;\frac{2}{1\cdot2} + \frac{2}{2\cdot3} + \frac{2}{3\cdot4} + \frac{2}{4\cdot5} + \hdots + \frac{2}{n(n+1)}$

. . . . . . . . . . $=\;2\left[\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \hdots + \frac{1}{n(n+1)}\right]$

. . . . . . . . . . $= \;2\left[\left(\frac{1}{1}-\frac{1}{2}\right) + \left(\frac{1}{2}-\frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \hdots + \left(\frac{1}{n} - \frac{1}{n+1}\right)\right]$

All the "middle" terms cancel out and we are left with:

. . . . . $S_n \;\;=\;\;2\left[1 - \frac{1}{n+1}\right] \;\;=\;\;\frac{2n}{n+1}$

For triangular number,
$T_n=\frac{n}{2}(n+1)$
$T_n +T_{n-1}=n^2$

Therefore, is it
$S_n= T_1+ T_2+ T_3+ \cdots +T_{n-1}+ T_n$
$S_n=\frac{n}{2}(n^2)$
?

9. $S_1=\frac{2(1)}{1+1}$

If $S_k=\frac{2k}{k+1}$ then

$S_{k+1}=\frac{2k}{k+1}+\frac{2}{(k+1)(k+2)}=\frac{ 2k(k+2)+2}{(k+1)(k+2)}=\frac{2(k^2+2k+1)}{(k+1)(k+ 2)}=\frac{2(k+1)}{k+2}$

10. Originally Posted by JaneBennet
$S_1=\frac{2(1)}{1+1}$

If $S_k=\frac{2k}{k+1}$ then

$S_{k+1}=\frac{2k}{k+1}+\frac{2}{(k+1)(k+2)}=\frac{ 2k(k+2)+2}{(k+1)(k+2)}=\frac{2(k^2+2k+1)}{(k+1)(k+ 2)}=\frac{2(k+1)}{k+2}$
I mean triangular number, 1,3,6,10,15,..., not the series 1st I had posted.