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This is the question:

Find the sum of $$\frac1 9+\frac2{9^2}+\frac3{9^3}+\ldots +\frac n{9^n}+\ldots$$

What I did was convert the whole equation to

$$\frac{1\cdot 9^{n-1}+2\cdot 9^{n-2}+3\cdot 9^{n-3}+\ldots +n}{9^n}$$

I have no clue about what to do next.

I would probably think that you would have to identify the sequence but I don't know how to get there from this.

4 Answers4

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Hint:

Use a letter instead of $1/9$, $x$ for instance; you get $$x+2x^2+3x^3+\dots+nx^n+\dotsm=x(1+2x+3x^2+\dots + nx^{n-1}+\dotsm)=x\, (\; ? \;)'$$

Bernard
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  • Sorry, but I still don't get it. How do I find the value of 1 + 2x + 3x^2 ...? –  Aug 24 '20 at 23:42
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    You have to find the value of its antiderivative first (in function of $x$), which is well-known. – Bernard Aug 24 '20 at 23:47
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No derivatives here, just geometric series:

$$\frac1 9+\frac2{9^2}+\frac3{9^3}+\cdots$$

$$=\frac19+\frac1{9^2}+\frac1{9^2}+\cdots$$

$$+\frac1{9^2}+\frac1{9^3}+\cdots$$

$$+\frac1{9^3}+\cdots$$

$$+\cdots$$

$$=\frac18$$

$$+\frac18\frac19$$

$$+\frac18\frac1{9^2}$$

$$+\cdots$$

$$= \frac18\left(1+\frac19+\frac1{9^2}+\cdots\right)=\frac18\left(\frac98\right)=\frac9{64}$$

J. W. Tanner
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  • Might I ask you where you got 1/8 from? –  Aug 25 '20 at 00:22
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    The first $\frac18$ is $\frac19+\frac1{9^2}+\frac1{9^3}+\cdots$. The $\frac18$ in front of $\large($ was factored out from each line – J. W. Tanner Aug 25 '20 at 00:23
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    Ok, thanks I was just confused about how you got the 1/8 but you cleared it up. –  Aug 25 '20 at 00:27
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HINT

We are looking for

$$\sum_{n=1}^\infty \frac{n}{9^n}=\sum_{n=1}^\infty nx^n$$

with $x=\frac19$ then we can use that

$$\frac d{dx}\sum_{n=0}^\infty x^n=\frac1x\sum_{n=1}^\infty nx^n$$

user
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hint

Consider the geometric series $$\sum (\frac{x}{9})^n$$

and its derivative

$$\sum \frac{n}{9^n}x^{n-1}$$

think about $ x=1$.

hamam_Abdallah
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