Question

. Evaluate the series 1 + 0.1 + 0.01 + . . .

Answer 1

We can employ a simple repeated decimal trick:

$x=1.111\ldots$

$0.1x=0.111\ldots$

$\implies x-0.1x=1\implies0.9x=1\implies x=\dfrac1{0.9}=\dfrac{10}9$

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Alternatively, we can compute the partial sum of the series.

$\displaystyle S_n=\sum_{k=0}^n\dfrac1{10^k}$

$S_n=1+0.1+0.01+\cdots+\dfrac1{10^n}$

$0.1S_n=0.1+0.01+0.001+\cdots+\dfrac1{10^{n+1}}$

$\implies S_n-0.1S_n=0.9S_n=1-\dfrac1{10^{n+1}}$

$\implies S_n=\dfrac{10}9-\dfrac9{10^n}$

As $n\to\infty$, the second term vanishes and we're left with $\dfrac{10}9$. Notice that this is really just a more formal version of the earlier trick.

Answer 2

Answer:

1.11

Step-by-step explanation: