Question

S= 1 + 1/1+2 + 1/1+2+3 + 1/1+2+3+4 + ... + n terms (telescopic series)

Answer 1

$\green{\large\underline{\sf{Solution-}}}$

Given series is

$\rm :\longmapsto\:S = 1 + \dfrac{1}{1 + 2} + \dfrac{1}{1 + 2 + 3} + - - n \: terms$

So, nth term of the series is

$\rm :\longmapsto\:T_n = \dfrac{1}{1 + 2 + 3 + 4 + - - - n \: terms}$

$\rm :\longmapsto\:T_n = \dfrac{1}{ \: \: \: \displaystyle\sum_{k=1}^n \: k \: \: \: }$

$\rm :\longmapsto\:T_n = \dfrac{1}{\dfrac{n(n + 1)}{2} }$

$\rm :\longmapsto\:T_n =2 \times \dfrac{1}{n(n + 1)}$

$\rm :\longmapsto\:T_n =2 \times \dfrac{n + 1 - n}{n(n + 1)}$

$\rm :\longmapsto\:T_n =2 \times \bigg[\dfrac{n + 1}{n(n + 1)} - \dfrac{n}{n(n + 1)} \bigg]$

$\rm :\longmapsto\:T_n =2 \bigg[\dfrac{1}{n} - \dfrac{1}{n + 1} \bigg]$

So, on substituting n = 1, 2, 3, - - - ,n, we get

$\rm :\longmapsto\:T_1 =2 \bigg[\dfrac{1}{1} - \dfrac{1}{2} \bigg]$

$\rm :\longmapsto\:T_2 =2 \bigg[\dfrac{1}{2} - \dfrac{1}{3} \bigg]$

$\rm :\longmapsto\:T_3 =2 \bigg[\dfrac{1}{3} - \dfrac{1}{4} \bigg]$

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$\rm :\longmapsto\:T_n =2 \bigg[\dfrac{1}{n} - \dfrac{1}{n + 1} \bigg]$

Now, we know that

$\rm :\longmapsto\:S = T_1 + T_2 + T_3 + - - - + T_n$

$\rm \:  =  \: 2\bigg[1 - \cancel \dfrac{1}{2} +\cancel \dfrac{1}{2} -\cancel \dfrac{1}{3} + - - - +\cancel \dfrac{1}{n} - \dfrac{1}{n + 1} \bigg]$

$\rm \:  =  \: 2\bigg[1 - \dfrac{1}{n + 1} \bigg]$

$\rm \:  =  \: 2\bigg[\dfrac{n + 1 - 1}{n + 1} \bigg]$

$\rm \:  =  \: 2\bigg[\dfrac{n}{n + 1} \bigg]$

$\rm \:  =  \: \dfrac{2n}{n + 1}$

Hence,

$\boxed{\tt{ S = 1+\dfrac{1}{1 + 2}+\dfrac{1}{1 + 2 + 3} + - - n \:terms = \dfrac{2n}{n + 1}}}$

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Additional Information :-

$\red{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\sum_{k=1}^n1 = n \: }}}$

$\red{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\sum_{k=1}^nk = \frac{n(n + 1)}{2} \: }}}$

$\red{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\sum_{k=1}^n {k}^{2} = \frac{n(n + 1)(2n + 1)}{6} \: }}}$

$\red{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\sum_{k=1}^n {k}^{3} = \bigg[\frac{n(n + 1)}{2}\bigg] ^{2} \: }}}$