# Find the sum of the expression below [duplicate]

Calculate: $$\dfrac{1}{2}+\dfrac{2}{2^²}+\dfrac{3}{2^3}+\dfrac{4}{2^4}+...=?$$

I try

$$\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+...\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+...\dfrac{1}{2^n}+\dfrac{1}{2^{n+1}}+\dfrac{1}{2^{ n+2}}$$

$$\dfrac{\dfrac{1}{2}}{1-\dfrac{1}{2}}+\dfrac{\dfrac{1}{4}}{1-\dfrac{1}{2}}+\dfrac{\dfrac{1}{8}}{1-\dfrac{1}{2}}+ ...\dfrac{\dfrac{1}{2^n}}{1-\dfrac{1}{2}}\\ 1+\dfrac{1}{2}+\dfrac{1}{4}+...\dfrac{2}{2^n}$$

how to finish

• Your sum can be rewritten as $\sum_{n=1}^\infty nx^n$ where $x=1/2$. Does this help? Commented May 12, 2023 at 12:09
• You are only one step from solving it yourselves, and I like that solution! The last sum is just $1+1/2+1/4+1/8+\ldots = 2$.
– user700480
Commented May 12, 2023 at 12:30

Let $$x=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+\ldots$$

Then, $$2x=1+\frac{2}{2}+\frac{3}{2^2}+\frac{4}{2^3}+\frac{5}{2^4}+\ldots$$

Then, $$2x-1-x=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\ldots=1$$

So, $$2x-1-x=1$$ i.e. $$x=2$$.

It is

$$\sum_n \dfrac{n}{2^n}$$

So take

$$\sum_n x^n=\dfrac{1}{1-x}$$

Derivating

$$\sum_n nx^{n-1}=\dfrac{1}{(1-x)^2}$$

$$x=1/2$$

$$\sum_n n(\dfrac{1}{2})^{n-1}$$

Adjust and you get the result.

• The OP may not know derivatives.
– user700480
Commented May 12, 2023 at 12:25
• What is the OP ?
– EDX
Commented May 12, 2023 at 12:27
• Original Poster
– user700480
Commented May 12, 2023 at 12:27