[tex] \large\underline{\sf{Solution-}}[/tex]
Given series is
[tex]\rm \longmapsto\:\displaystyle\sum_{k=1}^{\infty}\rm \displaystyle\sum_{n=1}^{\infty}\rm \frac{k}{ {2}^{n + k} } [/tex]
can be further rewritten as
[tex]\rm \: = \: \:\displaystyle\sum_{k=1}^{\infty}\rm \displaystyle\sum_{n=1}^{\infty}\rm \frac{k}{ {2}^{k} . {2}^{n} } [/tex]
can be further rewritten as
[tex]\rm \: = \: \displaystyle\sum_{k=1}^{\infty}\rm \bigg[\dfrac{k}{ {2}^{k}}\displaystyle\sum_{n=1}^{\infty}\rm \dfrac{1}{ {2}^{n} } \bigg][/tex]
[tex]\rm \: = \: \displaystyle\sum_{k=1}^{\infty}\rm \dfrac{k}{ {2}^{k} }\bigg(\dfrac{1}{2} + \dfrac{1}{ {2}^{2} } + \dfrac{1}{ {2}^{3} } - - - \infty \bigg) [/tex]
So, its an infinite GP series with
[tex] \purple{\rm \longmapsto\:a = \dfrac{1}{2}}[/tex]
[tex] \purple{\rm \longmapsto\:r = \dfrac{1}{2}}[/tex]
We know,
Sum of infinite GP series with common ratio r ( - 1 < r < 1 ) and first term a is given by
[tex]\boxed{\tt{ \: \: S_ \infty = \frac{a}{1 - r} , \: \: provided \: that \: |r| < 1}}[/tex]
So, 1using this, we get
[tex]\rm \: = \: \displaystyle\sum_{k=1}^{\infty}\rm \frac{k}{ {2}^{k} }\bigg( \frac{ \frac{1}{2} }{1 - \frac{1}{2} }\bigg)[/tex]
[tex]\rm \: = \: \displaystyle\sum_{k=1}^{\infty}\rm \frac{k}{ {2}^{k} }\bigg( \frac{ \frac{1}{2} }{ \frac{2 - 1}{2} }\bigg)[/tex]
[tex]\rm \: = \: \displaystyle\sum_{k=1}^{\infty}\rm \frac{k}{ {2}^{k} }\bigg( \frac{ \frac{1}{2} }{ \frac{1}{2} }\bigg)[/tex]
[tex]\rm \: = \: \displaystyle\sum_{k=1}^{\infty}\rm \frac{k}{ {2}^{k} }[/tex]
[tex]\rm \: = \: \dfrac{1}{2} + \dfrac{2}{ {2}^{2} } + \dfrac{3}{ {2}^{3} } + \dfrac{4}{ {2}^{4} } + - - - \infty [/tex]
Let assume that
[tex]\rm \longmapsto\:S_ \infty = \: \dfrac{1}{2} + \dfrac{2}{ {2}^{2} } + \dfrac{3}{ {2}^{3} } + \dfrac{4}{ {2}^{4} } + - - - \infty [/tex]
Now, its an infinite Arithmetico Geometrico Series, So multiply by 1/2 we get
[tex]\rm \longmapsto\:\dfrac{1}{2} S_ \infty = \: \dfrac{1}{ {2}^{2} } + \dfrac{2}{ {2}^{3} } + \dfrac{3}{ {2}^{4} } + \dfrac{4}{ {2}^{5} } + - - - \infty [/tex]
On Subtracting above two equations, we get
[tex]\rm \longmapsto\:\dfrac{1}{2} S_ \infty = \: \dfrac{1}{2} + \dfrac{1}{ {2}^{2} } + \dfrac{1}{ {2}^{3} } + \dfrac{1}{ {2}^{4} } + - - - \infty [/tex]
[tex]\rm \longmapsto\:\dfrac{1}{2} S_ \infty = \: \dfrac{ \dfrac{1}{2} }{1 - \dfrac{1}{2} } [/tex]
[tex]\rm \longmapsto\:\dfrac{1}{2} S_ \infty = \: \dfrac{ \dfrac{1}{2} }{ \dfrac{1}{2} } [/tex]
[tex]\rm \longmapsto\:\dfrac{1}{2} S_ \infty = \: 1[/tex]
[tex]\bf\implies \:S_ \infty = 2[/tex]
Hence,
[tex] \\ \rm \longmapsto\:\boxed{\tt{ \: \: \: \: \displaystyle\sum_{k=1}^{\infty}\rm \displaystyle\sum_{n=1}^{\infty}\rm \frac{k}{ {2}^{n + k} } = 2 \: \: \: \: }} \\ [/tex]
