Question

Express 1.191191119..... in summation notation
Please, help

Answer 1

@freckles

Answer 2

To get started, can you express this as a sum with good-ol'-fashioned "+" signs?

Answer 3

\(a_0= 10^0\\a_1=(10^1+9)*10^{-2}\\
a_2=(10^2+a_1)*10^{-3}\\\cdots\\
a_n=(10^n+a_{n-1})*10^{-(n+1)}\)

Answer 4

But I stuck there.

Answer 5

Sorry I was having connection issues.

Answer 6

So, if I add the LHS, I got 1.191191119..., but I don't know how to do with the rRHS

Answer 7

I don't know if recursion is particularly helpful here. Now I'm sure there are many ways to go about this, but here's one I see: \[1.191191191... = 119 \cdot 10^{-2} + 119 + \cdot 10^{-5} + 119 \cdot 10^{-8} +\cdots\]

Answer 8

That grouping is fairly easy to express with summation notation. Do you see how?

Answer 9

the third one is not that

Answer 10

Please clarify.

Answer 11

\(119*10^{-8}= 0.00000119\)