Sum-1-1-1-1-2-1-1-2-3-1-1-2-3-8016-

Question Number 79635 by TawaTawa last updated on 26/Jan/20

Sum : \frac{1}{1} + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + \dots + \frac{1}{1 + 2 + 3 + \dots + 8016}

Commented by mr W last updated on 26/Jan/20

= \underset{k = 1}{\sum^{n}} \frac{1}{1 + 2 + 3 + \dots + k}

= 2 \underset{k = 1}{\sum^{n}} \frac{1}{k (k + 1)}

= 2 \underset{k = 1}{\sum^{n}} (\frac{1}{k} - \frac{1}{k + 1})

= 2 (1 - \frac{1}{n + 1})

= \frac{2 n}{n + 1}

= \frac{2 \times 8016}{8017}

= \frac{16032}{8017}

i f n \to \infty

Σ \to 2

Commented by TawaTawa last updated on 26/Jan/20

God bless you sir

Commented by TawaTawa last updated on 26/Jan/20

I did not understand : 2 \underset{k = 1}{\sum^{n}} \frac{1}{k (k + 1)} sir

Commented by mr W last updated on 26/Jan/20

y o u {d o n}^{'} t k n o w 1 + 2 + 3 + \dots + k = ?

Commented by mr W last updated on 26/Jan/20

Commented by TawaTawa last updated on 26/Jan/20

ohh . i understand now sir .

1 + 2 + 3 + . . + n = \frac{n (n + 1)}{2} \Rightarrow \frac{1}{\frac{n (n + 1)}{2}} = \frac{2}{n (n + 1)}

so you take the 2 out .

Commented by TawaTawa last updated on 26/Jan/20

God bless you sir

Commented by john santu last updated on 27/Jan/20

1 + \frac{1}{3} + \frac{1}{6} + \frac{1}{10} + \frac{1}{15} + \frac{1}{21} + \dots = S_{m}

consider denumerator

1, 3, 6, 10, 15, 21, \dots

T_{n} = \frac{n^{2} + n}{2} = \frac{n (n + 1)}{2}

\underset{n = 1}{\sum^{8016}} \frac{2}{n (n + 1)} = 2 \underset{n = 1}{\sum^{8016}} \frac{1}{n} - \frac{1}{n + 1}

= 2 {\frac{1}{1} - \frac{1}{8017}} = \frac{16032}{8017}

Commented by TawaTawa last updated on 27/Jan/20

God bless you sir