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$$\lim _{n\to\infty}\sum_{k=0}^n\frac{k+1}{10^k}$$ What I've tried is to create a function out of it in order to derivate it, but I am getting nowhere. I would like to know if there is an easier way to it (I would like a high school level method for this, if there is one).

Parcly Taxel
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Jon9
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    I'd by splitting it into the sum of $k/10^k$ and $1/10^k$, where the latter is easily computable as it is a geometric series. – ViktorStein Jul 04 '19 at 16:36
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    Well, if you forget the "high-school level" part (or are willing to be a bit hazy on the justification of the steps) you can always consider the function $f$ defined for $x\in(-1,1)$ by $f(x) = \sum_{k=0}^\infty x^{k+1}$. You can compute a closed-form for it and differentiate that. You can also see that $f'(x) = \sum_{k=0}^\infty (k+1) x^k$, so what you want if $f'(1/10)$. – Clement C. Jul 04 '19 at 16:37

2 Answers2

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Write out the infinite sum: $$S=\frac1{10^0}+\frac2{10^1}+\frac3{10^2}+\frac4{10^3}+\dots$$ Divide by ten and subtract from $S$: $$\frac1{10}S=\frac1{10^1}+\frac2{10^2}+\frac3{10^3}+\frac4{10^4}+\dots$$ $$S-\frac1{10}S=\frac1{10^0}+\frac1{10^1}+\frac1{10^2}+\frac1{10^3}+\dots$$ This is a geometric series, whose sum can be easily calculated: $$\frac9{10}S=\frac1{1-1/10}=\frac{10}9$$ $$S=\frac{100}{81}$$

Parcly Taxel
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When $x|<1$ $$\frac{1}{1-x}=\sum_{k=0}^{\infty} x^k ~~~~(1).$$ Differentiate (1) w.r.t. $x$ to get $$\frac{1}{(1-x)^2}= \sum_{k=0}^{\infty} k x^{k-1} \Rightarrow \frac{x.}{(1-x)^2}= \sum_{k=0}^{\infty} k x^{k}~~~~(2)$$ Let us put $x=1/10$ in (1) and (2) and then add them to get $$\sum_{k=0}^{\infty} \frac{k+1}{10^k}=\frac{10}{9}+\frac{10}{81}=\frac{100}{81}.$$

Z Ahmed
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