The Easy Peasy Sum of Series Part 2

I hope that you will find my solution very easy.

Let $\sum _{ n=0 }^{ \infty }{ \frac { 2n\quad +\quad 1 }{ { 2 }^{ 2n+1 } } }$ = S

Therefore, S = $\frac { 1 }{ 2 } +\frac { 3 }{ { 2 }^{ 3 } } +\frac { 5 }{ { 2 }^{ 5 } } +\frac { 7 }{ { 2 }^{ 7 } } +\frac { 9 }{ { 2 }^{ 9 } } ......\infty$

Then, S/4 = $\frac { 1 }{ { 2 }^{ 3 } } +\frac { 3 }{ { 2 }^{ 5 } } +\frac { 5 }{ { 2 }^{ 7 } } +\frac { 7 }{ { 2 }^{ 9 } } ......\infty$

Subtract S/4 from S

S- S/4 = $\frac { 1 }{ 2 } +\frac { 2 }{ { 2 }^{ 3 } } +\frac { 2 }{ { 2 }^{ 5 } } +\frac { 2 }{ { 2 }^{ 7 } } +\frac { 2 }{ { 2 }^{ 9 } } ......\infty$

Now, we have an infinite geometric series with a = 1/4 and r = 1/4

Hence,

3S/4 = $\frac { 1 }{ 2 } + \frac { \frac { 1 }{ 4 } }{ 1\quad -\quad \frac { 1 }{ 4 } }$

3S/4 = $\frac { 1 }{ 2 } +\frac { 1 }{ 3 }$

So, S = 10/19 and therefore the answer $\boxed { 19 }$

Let us consider the sum:

$x+x^{3}+x^{5}+x^{7}+...\text{uptill infinite terms}$ where $|x|<1$

By the formula for the sum of the terms of an infinite Geometric Progression the above sum can be written as:

$\displaystyle\sum_{n=0}^{\infty} x^{2n+1}=\displaystyle\frac{x}{1-x^{2}}$

Now differentiating w.r.t. $x$ we get:

$\displaystyle\sum_{n=0}^{\infty} (2n+1)x^{2n}=\displaystyle\frac{1+x^{2}}{\left( 1-x^{2}\right)^{2}}$

Now substitute $x=\displaystyle\frac{1}{2}$ to obtain:

$\displaystyle\sum_{n=0}^{\infty} \frac{2n+1}{2^{2n}}=\displaystyle\frac{20}{9}$

Thus the required answer is $\displaystyle\frac{1}{2}\displaystyle\sum_{n=0}^{\infty} \frac{2n+1}{2^{2n}}=\boxed{\displaystyle\frac{10}{9}}$

$My\quad solution\quad is\quad somewhat\quad different.\\ Consider\quad 2n+1={ \frac { d }{ dx } { x }^{ 2n+1 }| }_{ x=1 },thus\quad \\ \sum _{ n=0 }^{ \infty }{ \frac { 2n+1 }{ { 2 }^{ 2n+1 } } } =\sum _{ n=0 }^{ \infty }{ \frac { d }{ dx } \frac { { x }^{ 2n+1 } }{ { 2 }^{ 2n+1 } } |_{ x=1 } } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =\frac { d }{ dx } \sum _{ n=0 }^{ \infty }{ { (\frac { { x }^{ 2 } }{ 4 } ) }^{ n } } \frac { x }{ 2 } |_{ x=1 }\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad ={ \frac { d }{ dx } }\frac { \frac { x }{ 2 } }{ 1-\frac { { x }^{ 2 } }{ 4 } } |_{ x=1 }\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =\frac { 8+2{ x }^{ 2 } }{ { (4-{ x }^{ 2 }) }^{ 2 } } |_{ x=1 }\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =\frac { 10 }{ 9 } .$

Here is a non-calculus and heavy summation notation approach: Note that $\frac{1}{2^2} = \frac{1}{2^2} \\ \frac{2}{2^4} = \frac{1}{2^4} + \frac{1}{2^4} \\ \frac{3}{2^6} = \frac{1}{2^6} + \frac{1}{2^6} + \frac{1}{2^6} \\ \vdots = \cdots$

Hence we can "sum vertically" to see that $\displaystyle\sum_{n=1}^{\infty} \frac{n}{2^{2n}} = \sum_{n=1}^{\infty} \left(\sum_{i = n}^{\infty} \frac{1}{2^{2i}} \right) \\ = \displaystyle\sum_{n=1}^{\infty} 2^{-2n} \left(\sum_{i=0}^{\infty} \frac{1}{2^{2i}} \right) = \sum_{n=1}^{\infty} 2^{-2n} \frac{1}{1- 1/4} \\ =\frac{4}{3} \displaystyle\sum_{n=1}^{\infty} 2^{-2n} = \frac{4}{3}\cdot \frac{1}{4} \cdot \frac{4}{3} = \frac{4}{9}$ .

Therefore

$\displaystyle\sum_{n=0}^{\infty} \frac{2n+1}{2^{2n+1}} = \sum_{n=1}^{\infty} \frac{n}{2^{2n}} + \sum_{n=0}^{\infty} \frac{1}{2^{2n+1}} = \frac{4}{9} + \frac{1}{2} \cdot \frac{4}{3} = \frac{10}{9}$

Hence $m/m = 10/9$ and $m + n = 10 + 9 = \mathbf{19}$ .

(NB: This method is not recommended, unless you are running out of time in the exam and wants to get the 1 mark for the answer mark, or if this is a fill-in-the-blank)

Let $S(N)$ be $\sum\limits^N_{n=0}\frac{2n+1}{2^{2n+1}}$ .

S(1)=0.5

S(2)=0.875

S(3)=1.03125

S(4)=1.0859375

S(5)=1.103515625

S(6)=1.10888671875

S(7)=1.11047363281

Enough evidence to know that it approaches to 1.111111... = 10/9

For $|x| < 1$ ,

$\begin{aligned} \sum_{k=0}^\infty x^{2k+1} & = \frac x{1-x^2} & \small \color{#3D99F6} \text{Differentiate both sides w.r.t. }x \\ \sum_{k=0}^\infty (2k+1)x^{2k} & = \frac {1+x^2}{(1-x^2)^2} & \small \color{#3D99F6} \text{Multiply both sides by }x \\ \sum_{k=0}^\infty (2k+1)x^{2k+1} & = \frac {x(1+x^2)}{(1-x^2)^2} & \small \color{#3D99F6} \text{Putting }x = \frac 12 \\ \sum_{k=0}^\infty \frac {2k+1}{2^{2k+1}} & = \frac {\frac 12\left(1+\frac 14\right)}{\left(1-\frac 14\right)^2} = \frac {10}9 \end{aligned}$

Therefore, $m+n=10+9 = \boxed{19}$ .